-34=-2z^2-32z

Solution for -34=-2z^2-32z equation:

-34=-2z^2-32z

We move all terms to the left:

-34-(-2z^2-32z)=0

We get rid of parentheses

2z^2+32z-34=0

a = 2; b = 32; c = -34;
Δ = b2-4ac
Δ = 322-4·2·(-34)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(32)-36}{2*2}=\frac{-68}{4}
=-17
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(32)+36}{2*2}=\frac{4}{4}
=1
$