140=(8+2x)(12+2x)

Solution for 140=(8+2x)(12+2x) equation:

140=(8+2x)(12+2x)

We move all terms to the left:

140-((8+2x)(12+2x))=0

We add all the numbers together, and all the variables

-((2x+8)(2x+12))+140=0

We multiply parentheses ..

-((+4x^2+24x+16x+96))+140=0


We calculate terms in parentheses: -((+4x^2+24x+16x+96)), so:

(+4x^2+24x+16x+96)

We get rid of parentheses

4x^2+24x+16x+96

We add all the numbers together, and all the variables

4x^2+40x+96

Back to the equation:

-(4x^2+40x+96)


We get rid of parentheses

-4x^2-40x-96+140=0

We add all the numbers together, and all the variables

-4x^2-40x+44=0

a = -4; b = -40; c = +44;
Δ = b2-4ac
Δ = -402-4·(-4)·44
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-48}{2*-4}=\frac{-8}{-8}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+48}{2*-4}=\frac{88}{-8}
=-11
$