Answer:
[tex] x = -2 [/tex] or [tex] x= -3 [/tex]
The solution is real.
Step-by-step explanation:
x^2 + 5x + 6 = 0
[tex] x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex]
[tex] x = \dfrac{-5 \pm \sqrt{5^2 - 4(1)(6)}}{2(1)} [/tex]
[tex] x = \dfrac{-5 \pm \sqrt{25 - 24}}{2} [/tex]
[tex] x = \dfrac{-5 \pm \sqrt{1}}{2} [/tex]
[tex] x = \dfrac{-4}{2} [/tex] or [tex] x = \dfrac{-6}{2} [/tex]
[tex] x = -2 [/tex] or [tex] x= -3 [/tex]
The solution is real.
