(8+2x)(10+2x)=104

Solution for (8+2x)(10+2x)=104 equation:

(8+2x)(10+2x)=104

We move all terms to the left:

(8+2x)(10+2x)-(104)=0

We add all the numbers together, and all the variables

(2x+8)(2x+10)-104=0

We multiply parentheses ..

(+4x^2+20x+16x+80)-104=0

We get rid of parentheses

4x^2+20x+16x+80-104=0

We add all the numbers together, and all the variables

4x^2+36x-24=0

a = 4; b = 36; c = -24;
Δ = b2-4ac
Δ = 362-4·4·(-24)
Δ = 1680
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1680}=\sqrt{16*105}=\sqrt{16}*\sqrt{105}=4\sqrt{105}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(36)-4\sqrt{105}}{2*4}=\frac{-36-4\sqrt{105}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(36)+4\sqrt{105}}{2*4}=\frac{-36+4\sqrt{105}}{8}
$