(20+2x)(28+2x)=1008

Solution for (20+2x)(28+2x)=1008 equation:

(20+2x)(28+2x)=1008

We move all terms to the left:

(20+2x)(28+2x)-(1008)=0

We add all the numbers together, and all the variables

(2x+20)(2x+28)-1008=0

We multiply parentheses ..

(+4x^2+56x+40x+560)-1008=0

We get rid of parentheses

4x^2+56x+40x+560-1008=0

We add all the numbers together, and all the variables

4x^2+96x-448=0

a = 4; b = 96; c = -448;
Δ = b2-4ac
Δ = 962-4·4·(-448)
Δ = 16384
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16384}=128$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-128}{2*4}=\frac{-224}{8}
=-28
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+128}{2*4}=\frac{32}{8}
=4
$