165=(8+2x)(4+2x)

Solution for 165=(8+2x)(4+2x) equation:

165=(8+2x)(4+2x)

We move all terms to the left:

165-((8+2x)(4+2x))=0

We add all the numbers together, and all the variables

-((2x+8)(2x+4))+165=0

We multiply parentheses ..

-((+4x^2+8x+16x+32))+165=0


We calculate terms in parentheses: -((+4x^2+8x+16x+32)), so:

(+4x^2+8x+16x+32)

We get rid of parentheses

4x^2+8x+16x+32

We add all the numbers together, and all the variables

4x^2+24x+32

Back to the equation:

-(4x^2+24x+32)


We get rid of parentheses

-4x^2-24x-32+165=0

We add all the numbers together, and all the variables

-4x^2-24x+133=0

a = -4; b = -24; c = +133;
Δ = b2-4ac
Δ = -242-4·(-4)·133
Δ = 2704
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2704}=52$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-24)-52}{2*-4}=\frac{-28}{-8}
=3+1/2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-24)+52}{2*-4}=\frac{76}{-8}
=-9+1/2
$