Answer
The solutions are $x=-2\pm\dfrac{\sqrt{6}}{3}i$
Work Step by Step
$3x^{2}+12x=-14$
Take out common factor $3$ from the left side:
$3(x^{2}+4x)=-14$
Take $3$ to divide the right side:
$x^{2}+4x=-\dfrac{14}{3}$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this particular case, $b=4$:
$x^{2}+4x+\Big(\dfrac{4}{2}\Big)^{2}=-\dfrac{14}{3}+\Big(\dfrac{4}{2}\Big)^{2}$
$x^{2}+4x+4=-\dfrac{14}{3}+4$
$x^{2}+4x+4=-\dfrac{2}{3}$
Factor the left side of the equation, which is a perfect square trinomial:
$(x+2)^{2}=-\dfrac{2}{3}$
Take the square root of both sides:
$\sqrt{(x+2)^{2}}=\pm\sqrt{-\dfrac{2}{3}}$
$x+2=\pm\sqrt{\dfrac{2}{3}}i$
$x+2=\pm\dfrac{\sqrt{6}}{3}i$
Solve for $x$:
$x=-2\pm\dfrac{\sqrt{6}}{3}i$
