(20+2x)(28+2x)=1584

Solution for (20+2x)(28+2x)=1584 equation:

(20+2x)(28+2x)=1584

We move all terms to the left:

(20+2x)(28+2x)-(1584)=0

We add all the numbers together, and all the variables

(2x+20)(2x+28)-1584=0

We multiply parentheses ..

(+4x^2+56x+40x+560)-1584=0

We get rid of parentheses

4x^2+56x+40x+560-1584=0

We add all the numbers together, and all the variables

4x^2+96x-1024=0

a = 4; b = 96; c = -1024;
Δ = b2-4ac
Δ = 962-4·4·(-1024)
Δ = 25600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25600}=160$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-160}{2*4}=\frac{-256}{8}
=-32
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+160}{2*4}=\frac{64}{8}
=8
$