(8+2x)(3+2x)=120

Solution for (8+2x)(3+2x)=120 equation:

(8+2x)(3+2x)=120

We move all terms to the left:

(8+2x)(3+2x)-(120)=0

We add all the numbers together, and all the variables

(2x+8)(2x+3)-120=0

We multiply parentheses ..

(+4x^2+6x+16x+24)-120=0

We get rid of parentheses

4x^2+6x+16x+24-120=0

We add all the numbers together, and all the variables

4x^2+22x-96=0

a = 4; b = 22; c = -96;
Δ = b2-4ac
Δ = 222-4·4·(-96)
Δ = 2020
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:

$\sqrt{\Delta}=\sqrt{2020}=\sqrt{4*505}=\sqrt{4}*\sqrt{505}=2\sqrt{505}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2\sqrt{505}}{2*4}=\frac{-22-2\sqrt{505}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2\sqrt{505}}{2*4}=\frac{-22+2\sqrt{505}}{8}
$