• \({\Delta=b^2-4ac}\)
• \({a=3}\)
• \({b=-21}\)
• \({c=36}\)
• \({\Delta=(-21)^2-4 \cdot 3 \cdot 36 }\)

\({\textcolor{white}{\Delta}=441-432 }\)

\({\textcolor{white}{\Delta}=9 }\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -(-21) \pm \sqrt{9}}{\displaystyle 2 \cdot 3 }}\)




\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle 21 \pm 3}{\displaystyle 6 }}\)

pentru că \({- \cdot -=+}\)



• \({x_1=\frac{\displaystyle 21 + 3}{\displaystyle 6}}\)



\({\textcolor{white}{x_1}=\frac{\displaystyle 24}{\displaystyle 6}}\)

\({\textcolor{white}{x_1}=4}\)



• \({x_2=\frac{\displaystyle 21 - 3}{\displaystyle 6}}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle 18}{\displaystyle 6}}\)

\({\textcolor{white}{x_2}=3}\)

• dăm factor comun pe 3 și obținem \({E(x)=3(x^2-7x+12)}\)
• calculăm soluțiile ecuației \({x^2-7x+12=0}\)
• \({\Delta=(-7)^2-4 \cdot 1 \cdot 12 }\)

\({\textcolor{white}{\Delta}=49-48 }\)

\({\textcolor{white}{\Delta}=1 }\)

• \({x_{1,2}=\frac{\displaystyle -(-7) \pm \sqrt{1}}{\displaystyle 2 \cdot 1}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle 7 \pm 1}{\displaystyle 2 }}\)

• \({x_{1}=\frac{\displaystyle 7+1 }{\displaystyle 2 }}\)



\({x_{1}=4}\)

• \({x_{2}=\frac{\displaystyle 7-1 }{\displaystyle 2 }}\)



\({x_{2}=3}\)

• rezultă că \({x^2-7x+12=(x-4)(x-3)}\)
• obținem că \({E(x)=3(x-4)(x-3)}\)

• \({\Delta=b^2-4ac}\)
• \({a=2}\)
• \({b=9}\)
• \({c=-5}\)
• \({\Delta=9^2-4 \cdot 2 \cdot (-5) }\)

\({\textcolor{white}{\Delta}=81+40 }\)

\({\textcolor{white}{\Delta}=121 }\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -9 \pm \sqrt{121}}{\displaystyle 2 \cdot 2 }}\)




\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -9 \pm 11}{\displaystyle 4 }}\)



• \({x_1=\frac{\displaystyle -9 + 11}{\displaystyle 4}}\)



\({\textcolor{white}{x_1}=\frac{\displaystyle 2}{\displaystyle 4}}\)

\({\textcolor{white}{x_1}=\frac{\displaystyle 1}{\displaystyle 2}}\)



• \({x_2=\frac{\displaystyle -9 - 11}{\displaystyle 4}}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle -20}{\displaystyle 4}}\)

\({\textcolor{white}{x_2}=-5}\)

\({\textcolor{white}{2x^2+9x-5}=2(x-\frac{\displaystyle 1}{\displaystyle 2})(x+5)}\)

\({\textcolor{white}{2x^2+9x-5}=(2 \cdot x- \cancel {2} \cdot \frac{\displaystyle 1}{\displaystyle \cancel 2})(x+5)}\)

\({\textcolor{white}{2x^2+9x-5}=(2x-1)(x+5)}\)

• \({\Delta=b^2-4ac}\)
• \({a=1}\)
• \({b=5}\)
• \({c=6}\)
• \({\Delta=5^2-4 \cdot 1 \cdot 6 }\)

\({\textcolor{white}{\Delta}=25-24 }\)

\({\textcolor{white}{\Delta}=1 }\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -5 \pm \sqrt{1}}{\displaystyle 2 \cdot 1 }}\)




\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -5 \pm 1}{\displaystyle 2 }}\)



• \({x_1=\frac{\displaystyle -5+1}{\displaystyle 2}}\)



\({\textcolor{white}{x_1}=\frac{\displaystyle -4}{\displaystyle 2}}\)

\({\textcolor{white}{x_1}=-2}\)



• \({x_2=\frac{\displaystyle -5 - 1}{\displaystyle 2}}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle -6}{\displaystyle 2}}\)

\({\textcolor{white}{x_2}=-3}\)

\({\textcolor{white}{x^2+5x+6}=(x+2)(x+3)}\)

pentru că \({- \cdot -=+}\)

• \({\Delta=b^2-4ac}\)
• \({a=1}\)
• \({b=-1}\)
• \({c=-6}\)
• \({\Delta=(-1)^2-4 \cdot 1 \cdot (-6) }\)

\({\textcolor{white}{\Delta}=1+24 }\)

\({\textcolor{white}{\Delta}=25 }\)

pentru că \({- \cdot -=+}\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -(-1) \pm \sqrt{25}}{\displaystyle 2 \cdot 1 }}\)




\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle 1 \pm 5}{\displaystyle 2 }}\)



• \({x_1=\frac{\displaystyle 1+5}{\displaystyle 2}}\)

\({\textcolor{white}{x_1}=\frac{\displaystyle 6}{\displaystyle 2}}\)

\({\textcolor{white}{x_1}=3}\)



• \({x_2=\frac{\displaystyle 1-5}{\displaystyle 2}}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle -4}{\displaystyle 2}}\)

\({\textcolor{white}{x_2}=-2}\)

\({\textcolor{white}{x^2-x-6}=(x-3)(x+2)}\)

pentru că \({- \cdot -=+}\)

• \({\Delta=b^2-4ac}\)
• \({a=4}\)
• \({b=16}\)
• \({c=-20}\)
• \({\Delta=(16)^2-4 \cdot 4 \cdot (-20) }\)

\({\textcolor{white}{\Delta}=256+320 }\)

\({\textcolor{white}{\Delta}=576 }\)

pentru că \({- \cdot -=+}\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -16 \pm \sqrt{576}}{\displaystyle 2 \cdot 4 }}\)




\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -16 \pm 24}{\displaystyle 8 }}\)



• \({x_1=\frac{\displaystyle -16 + 24}{\displaystyle 8}}\)



\({\textcolor{white}{x_1}=\frac{\displaystyle 8}{\displaystyle 8}}\)

\({\textcolor{white}{x_1}=1}\)



• \({x_2=\frac{\displaystyle -16 - 24}{\displaystyle 8}}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle -40}{\displaystyle 8}}\)

\({\textcolor{white}{x_2}=-5}\)

\({\textcolor{white}{4x^2+16x-20}=4(x-1)(x+5)}\)

pentru că \({- \cdot -=+}\)

• dăm factor comun pe 4 și obținem \({E(x)=4(x^2+4x-5)}\)
• calculăm soluțiile ecuației \({x^2+4x-5=0}\)
• \({\Delta=4^2-4 \cdot 1 \cdot (-5) }\)

\({\textcolor{white}{\Delta}=16+20 }\)

\({\textcolor{white}{\Delta}=36 }\)

• \({x_{1,2}=\frac{\displaystyle -4 \pm \sqrt{36}}{\displaystyle 2 \cdot 1}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -4 \pm 6}{\displaystyle 2 }}\)

• \({x_{1}=\frac{\displaystyle -4+6 }{\displaystyle 2 }}\)



\({\textcolor{white}{x_{1}}=\frac{\displaystyle 2 }{\displaystyle 2 }}\)

\({x_{1}=1}\)

• \({x_{2}=\frac{\displaystyle -4-6 }{\displaystyle 2 }}\)



\({\textcolor{white}{x_{2}}=\frac{\displaystyle -10 }{\displaystyle 2 }}\)

\({x_{2}=-5}\)

• rezultă că \({x^2+4x-5=(x-1)[x-(-5)]}\)

rezultă că \({\textcolor{white}{x^2+4x-5}=(x-1)(x+5)}\)

• obținem că \({E(x)=4(x-1)(x+5)}\)

pentru că \({- \cdot -=+}\)

• \({\Delta=b^2-4ac}\)
• \({a=1}\)
• \({b=\frac{\displaystyle 1}{\displaystyle 3}}\)
• \({c=-\frac{\displaystyle 7}{\displaystyle 4}}\)
• \({\Delta=\left(\frac{\displaystyle 1}{\displaystyle 3}\right)^2-4 \cdot 1 \cdot \left(-\frac{\displaystyle 7}{\displaystyle 4} \right) }\)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 9}+7 }\)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 9}+\frac{\displaystyle 7}{\displaystyle 1} }\) (vom aduce la același numitor)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 9}+\frac{\displaystyle 7 \cdot 9}{\displaystyle 1 \cdot 9} }\) (amplificăm cu 9)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 9}+\frac{\displaystyle 63}{\displaystyle 9} }\)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 64}{\displaystyle 9} }\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 3} \pm \sqrt{\frac{\displaystyle 64}{\displaystyle 9}}}{\displaystyle 2 \cdot 1 }}\)




\({\textcolor{white}{x_{1,2}}= \frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 3} \pm \frac{\displaystyle 8}{\displaystyle 3}}{\displaystyle 2 }}\)



• \({x_1=\frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 3} + \frac{\displaystyle 8}{\displaystyle 3}}{\displaystyle 2 }}\)



\({\textcolor{white}{x_1}=\frac{\displaystyle \frac{\displaystyle 7}{\displaystyle 3}}{\frac{\displaystyle 2}{\displaystyle 1} }}\)

\({\textcolor{white}{x_1}=\frac{\displaystyle 7}{\displaystyle 3} \cdot \frac{\displaystyle 1}{\displaystyle 2}}\)

\({x_1=\frac{\displaystyle 7}{\displaystyle 6} }\)

pentru că \({\frac{\displaystyle \frac{\displaystyle a}{\displaystyle b}}{\frac{\displaystyle \textcolor{red}{c}}{\displaystyle \textcolor{red}{d}} }=\frac{\displaystyle a}{\displaystyle b} \cdot \frac{\displaystyle \textcolor{red}{d}}{\displaystyle \textcolor{red}{c}}}\)



• \({x_2=\frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 3} - \frac{\displaystyle 8}{\displaystyle 3}}{\displaystyle 2 }}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle -\frac{\displaystyle 9}{\displaystyle 3}}{\displaystyle 2 }}\)

\({\textcolor{white}{x_2}=-\frac{\displaystyle 3}{\displaystyle 2} }\)

\({\textcolor{white}{x^2+\frac{\displaystyle 1}{\displaystyle 3}x-\frac{\displaystyle 7}{\displaystyle 4}}=\left(x-\frac{\displaystyle 7}{\displaystyle 6}\right)\left(x+\frac{\displaystyle 3}{\displaystyle 2}\right)}\)

pentru că \({- \cdot -=+}\)

• \({\Delta=b^2-4ac}\)
• \({a=1}\)
• \({b=\frac{\displaystyle 1}{\displaystyle 5}}\)
• \({c=-\frac{\displaystyle 4}{\displaystyle 5}}\)
• \({\Delta=\left(\frac{\displaystyle 1}{\displaystyle 5}\right)^2-4 \cdot 1 \cdot \left(-\frac{\displaystyle 4}{\displaystyle 5} \right) }\)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 25}+\frac{\displaystyle 16}{\displaystyle 5} }\)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 25}+\frac{\displaystyle 16 \cdot 5}{\displaystyle 5 \cdot 5} }\) (amplificăm cu 5)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 1}{\displaystyle 25} +\frac{\displaystyle 80}{\displaystyle 25}}\)

\({\textcolor{white}{\Delta}=\frac{\displaystyle 81}{\displaystyle 25}}\)

• \({x_{1,2}=\frac{\displaystyle -b \pm \sqrt{\Delta}}{\displaystyle 2a}}\)



\({\textcolor{white}{x_{1,2}}=\frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 5} \pm \sqrt{\frac{\displaystyle 81}{\displaystyle 25}}}{\displaystyle 2 \cdot 1 }}\)




\({\textcolor{white}{x_{1,2}}= \frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 5} \pm \frac{\displaystyle 9}{\displaystyle 5}}{\displaystyle 2 }}\)



• \({x_1=\frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 5} + \frac{\displaystyle 9}{\displaystyle 5}}{\displaystyle 2 }}\)



\({\textcolor{white}{x_1}=\frac{\displaystyle \frac{\displaystyle 8}{\displaystyle 5}}{\frac{\displaystyle 2}{\displaystyle 1} }}\)

\({\textcolor{white}{x_1}=\frac{\displaystyle \cancel {8}^4}{\displaystyle 5} \cdot \frac{\displaystyle 1}{\displaystyle \cancel {2}_1}}\)

\({x_1=\frac{\displaystyle 4}{\displaystyle 5} }\)

pentru că \({\frac{\displaystyle \frac{\displaystyle a}{\displaystyle b}}{\frac{\displaystyle \textcolor{red}{c}}{\displaystyle \textcolor{red}{d}} }=\frac{\displaystyle a}{\displaystyle b} \cdot \frac{\displaystyle \textcolor{red}{d}}{\displaystyle \textcolor{red}{c}}}\)



• \({x_2=\frac{\displaystyle -\frac{\displaystyle 1}{\displaystyle 5} - \frac{\displaystyle 9}{\displaystyle 5}}{\displaystyle 2 }}\)



\({\textcolor{white}{x_2}=\frac{\displaystyle -\frac{\displaystyle 10}{\displaystyle 5}}{\displaystyle 2 }}\)

\({\textcolor{white}{x_2}=\frac{\displaystyle -2}{\displaystyle 2 }}\)

\({\textcolor{white}{x_2}=-1 }\)

\({\textcolor{white}{x^2+\frac{\displaystyle 1}{\displaystyle 5}x-\frac{\displaystyle 4}{\displaystyle 5}}=\left(x-\frac{\displaystyle 4}{\displaystyle 5}\right)(x+1)}\)

pentru că \({- \cdot -=+}\)