(d+4)(d-2)=55

Solution for (d+4)(d-2)=55 equation:

(d+4)(d-2)=55

We move all terms to the left:

(d+4)(d-2)-(55)=0

We multiply parentheses ..

(+d^2-2d+4d-8)-55=0

We get rid of parentheses

d^2-2d+4d-8-55=0

We add all the numbers together, and all the variables

d^2+2d-63=0

a = 1; b = 2; c = -63;
Δ = b2-4ac
Δ = 22-4·1·(-63)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$d_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-16}{2*1}=\frac{-18}{2}
=-9
$
$d_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+16}{2*1}=\frac{14}{2}
=7
$