0=2d^2+20d+32

Solution for 0=2d^2+20d+32 equation:

0=2d^2+20d+32

We move all terms to the left:

0-(2d^2+20d+32)=0

We add all the numbers together, and all the variables

-(2d^2+20d+32)=0

We get rid of parentheses

-2d^2-20d-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:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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