(24+2x)(20+2x)=320

Solution for (24+2x)(20+2x)=320 equation:

(24+2x)(20+2x)=320

We move all terms to the left:

(24+2x)(20+2x)-(320)=0

We add all the numbers together, and all the variables

(2x+24)(2x+20)-320=0

We multiply parentheses ..

(+4x^2+40x+48x+480)-320=0

We get rid of parentheses

4x^2+40x+48x+480-320=0

We add all the numbers together, and all the variables

4x^2+88x+160=0

a = 4; b = 88; c = +160;
Δ = b2-4ac
Δ = 882-4·4·160
Δ = 5184
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{5184}=72$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(88)-72}{2*4}=\frac{-160}{8}
=-20
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(88)+72}{2*4}=\frac{-16}{8}
=-2
$