Answer
$\displaystyle \frac{1}{2}$ and $-4$
Work Step by Step
Set the $RHS$ to 0 by subtracting 4 from both sides
$z(2z+7)=4$ $\qquad.../-2$
$2z^{2}+7z-4=0$
To factor the trinomial, we search for
factors of $2\times(-4)=-8, $ whose sum is $+7:\qquad $
We find $+8$ and $-1$.
$2z^{2}+7z-4=2z^{2}-z+8z-4=$
... and factor in pairs:
$=z(2z-1)+4(2z-1)$
$=(2z-1)(z+4)$
$(2z-1)(z+4)=0$
By the zero product principle,
$2z-1=0$ or $z+4=0$
$z=\displaystyle \frac{1}{2}$ or $z=-4$
The solutions are $\displaystyle \frac{1}{2}$ and $-4$.
