48=(12+2x)(20+2x)

Solution for 48=(12+2x)(20+2x) equation:

48=(12+2x)(20+2x)

We move all terms to the left:

48-((12+2x)(20+2x))=0

We add all the numbers together, and all the variables

-((2x+12)(2x+20))+48=0

We multiply parentheses ..

-((+4x^2+40x+24x+240))+48=0


We calculate terms in parentheses: -((+4x^2+40x+24x+240)), so:

(+4x^2+40x+24x+240)

We get rid of parentheses

4x^2+40x+24x+240

We add all the numbers together, and all the variables

4x^2+64x+240

Back to the equation:

-(4x^2+64x+240)


We get rid of parentheses

-4x^2-64x-240+48=0

We add all the numbers together, and all the variables

-4x^2-64x-192=0

a = -4; b = -64; c = -192;
Δ = b2-4ac
Δ = -642-4·(-4)·(-192)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-64)-32}{2*-4}=\frac{32}{-8}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-64)+32}{2*-4}=\frac{96}{-8}
=-12
$