Answer
$z=1$ and $z=-4$
Work Step by Step
$z^{2}+3z-4=0$
Take the $4$ to the right side of the equation:
$z^{2}+3z=4$
Add $\Big(\dfrac{b}{2}\Big)^{2}$ to both sides of the equation. In this case, $b=3$
$z^{2}+3z+\Big(\dfrac{3}{2}\Big)^{2}=4+\Big(\dfrac{3}{2}\Big)^{2}$
$z^{2}+3z+\dfrac{9}{4}=\dfrac{25}{4}$
Factor the expression on the left side of the equation, which is a perfect square trinomial:
$\Big(z+\dfrac{3}{2}\Big)^{2}=\dfrac{25}{4}$
Take the square root of both sides of the equation:
$\sqrt{\Big(z+\dfrac{3}{2}\Big)^{2}}=\sqrt{\dfrac{25}{4}}$
$z+\dfrac{3}{2}=\pm\dfrac{5}{2}$
Solve for $z$:
$z=\dfrac{-3\pm5}{2}$
The two solutions are:
$z=\dfrac{-3+5}{2}=1$
$z=\dfrac{-3-5}{2}=-4$
