320+76=(20+2x)(16+2x)

Solution for 320+76=(20+2x)(16+2x) equation:

320+76=(20+2x)(16+2x)

We move all terms to the left:

320+76-((20+2x)(16+2x))=0

We add all the numbers together, and all the variables

-((2x+20)(2x+16))+320+76=0

We add all the numbers together, and all the variables

-((2x+20)(2x+16))+396=0

We multiply parentheses ..

-((+4x^2+32x+40x+320))+396=0


We calculate terms in parentheses: -((+4x^2+32x+40x+320)), so:

(+4x^2+32x+40x+320)

We get rid of parentheses

4x^2+32x+40x+320

We add all the numbers together, and all the variables

4x^2+72x+320

Back to the equation:

-(4x^2+72x+320)


We get rid of parentheses

-4x^2-72x-320+396=0

We add all the numbers together, and all the variables

-4x^2-72x+76=0

a = -4; b = -72; c = +76;
Δ = b2-4ac
Δ = -722-4·(-4)·76
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-72)-80}{2*-4}=\frac{-8}{-8}
=1
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-72)+80}{2*-4}=\frac{152}{-8}
=-19
$