2z2+20z+32=0

Solution for 2z2+20z+32=0 equation:

2z^2+20z+32=0

a = 2; b = 20; c = +32;
Δ = b2-4ac
Δ = 202-4·2·32
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-12}{2*2}=\frac{-32}{4}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+12}{2*2}=\frac{-8}{4}
=-2
$