Answer
$x=-5\pm\sqrt{3}$
Work Step by Step
$x^{2}+10x+28=0$
Take the $28$ to the right side of the equation:
$x^{2}+10x=-28$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=10$
$x^{2}+10x+\Big(\dfrac{10}{2}\Big)^{2}=-28+\Big(\dfrac{10}{2}\Big)^{2}$
$x^{2}+10x+25=-28+25$
$x^{2}+10x+25=-3$
Factor the expression on the left side of the equation, which is a perfect square trinomial:
$(x+5)^{2}=-3$
Take the square root of both sides of the equation:
$\sqrt{(x+5)^{2}}=\sqrt{-3}$
$x+5=\pm\sqrt{3}i$
Solve for $x$:
$x=-5\pm\sqrt{3}$
