Answer
$x = -4~~$ or $~~x=-3$
Work Step by Step
A quadratic equation can be written in this form:
$ax^2 + bx+c = 0$
where $a,b,$ and $c$ are real numbers and $a \neq 0$
We can determine the values of $a, b,$ and $c$:
$x^2+7x+12 = 0$
$a = 1$
$b = 7$
$c=12$
We can use the quadratic formula to find the solutions of the equation:
$x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$
$x = \frac{-(7) \pm \sqrt{(7)^2-(4)(1)(12)}}{(2)(1)}$
$x = \frac{-7 \pm \sqrt{49-48}}{2}$
$x = \frac{-7 \pm \sqrt{1}}{2}$
$x = \frac{-7 \pm 1}{2}$
$x = \frac{-7 - 1}{2}~~$ or $~~x = \frac{-7 + 1}{2}$
$x = -4~~$ or $~~x=-3$
