(-4-2d)(4+2d)=-20

Solution for (-4-2d)(4+2d)=-20 equation:

(-4-2d)(4+2d)=-20

We move all terms to the left:

(-4-2d)(4+2d)-(-20)=0

We add all the numbers together, and all the variables

(-2d-4)(2d+4)-(-20)=0

We add all the numbers together, and all the variables

(-2d-4)(2d+4)+20=0

We multiply parentheses ..

(-4d^2-8d-8d-16)+20=0

We get rid of parentheses

-4d^2-8d-8d-16+20=0

We add all the numbers together, and all the variables

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

a = -4; b = -16; c = +4;
Δ = b2-4ac
Δ = -162-4·(-4)·4
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-8\sqrt{5}}{2*-4}=\frac{16-8\sqrt{5}}{-8}
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+8\sqrt{5}}{2*-4}=\frac{16+8\sqrt{5}}{-8}
$