x²+x-6=0 ⇔ x² + 3x - 2x - 6 = 0 ⇔ x(x+3) -2(x+3) =0 ⇔ (x+3)(x-2)=0
x+3=0 ⇒ x = -3
x-2=0 ⇒ x=2
Deci, ecuatia admite doua solutii : x = -3 si x = 2.
[tex]\it Sau:[/tex]
[tex]\it x^2+x-6 =0,\ a=1,\ b=1,\ c=-6[/tex]
[tex]\it\Delta =b^2-4ac =1^2-4\cdot1\cdot(-6) = 1+24=25[/tex]
[tex]\it x_{1,2} =\dfrac {-b \pm \sqrt{\Delta}}{2a} =\dfrac{-1\pm\sqrt{25}}{2\cdot1} =\dfrac{-1\pm5}{2}[/tex]
[tex]\begin{cases}\it x_1=\dfrac{-1-5}{2}=\dfrac{-6}{2}=-3
\\\;\\
\it x_2 = \dfrac{-1+5}{2}=\dfrac{4}{2}=2\end{cases}[/tex]
