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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

4x^2+24x+32

Back to the equation:

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


We add all the numbers together, and all the variables

x-(4x^2+24x+32)=0

We get rid of parentheses

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

We add all the numbers together, and all the variables

-4x^2-23x-32=0

a = -4; b = -23; c = -32;
Δ = b2-4ac
Δ = -232-4·(-4)·(-32)
Δ = 17
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{17}}{2*-4}=\frac{23-\sqrt{17}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{17}}{2*-4}=\frac{23+\sqrt{17}}{-8}
$