(d+4)(d-2)=16

Solution for (d+4)(d-2)=16 equation:

(d+4)(d-2)=16

We move all terms to the left:

(d+4)(d-2)-(16)=0

We multiply parentheses ..

(+d^2-2d+4d-8)-16=0

We get rid of parentheses

d^2-2d+4d-8-16=0

We add all the numbers together, and all the variables

d^2+2d-24=0

a = 1; b = 2; c = -24;
Δ = b2-4ac
Δ = 22-4·1·(-24)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-10}{2*1}=\frac{-12}{2}
=-6
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+10}{2*1}=\frac{8}{2}
=4
$