(20+2x)(24+2x)=1440

Solution for (20+2x)(24+2x)=1440 equation:

(20+2x)(24+2x)=1440

We move all terms to the left:

(20+2x)(24+2x)-(1440)=0

We add all the numbers together, and all the variables

(2x+20)(2x+24)-1440=0

We multiply parentheses ..

(+4x^2+48x+40x+480)-1440=0

We get rid of parentheses

4x^2+48x+40x+480-1440=0

We add all the numbers together, and all the variables

4x^2+88x-960=0

a = 4; b = 88; c = -960;
Δ = b2-4ac
Δ = 882-4·4·(-960)
Δ = 23104
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{23104}=152$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(88)-152}{2*4}=\frac{-240}{8}
=-30
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(88)+152}{2*4}=\frac{64}{8}
=8
$