(d-4)(d+2)=0

Solution for (d-4)(d+2)=0 equation:

(d-4)(d+2)=0

We multiply parentheses ..

(+d^2+2d-4d-8)=0

We get rid of parentheses

d^2+2d-4d-8=0

We add all the numbers together, and all the variables

d^2-2d-8=0

a = 1; b = -2; c = -8;
Δ = b2-4ac
Δ = -22-4·1·(-8)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*1}=\frac{-4}{2}
=-2
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*1}=\frac{8}{2}
=4
$