2x(x-8)+4x=20

Solution for 2x(x-8)+4x=20 equation:

2x(x-8)+4x=20

We move all terms to the left:

2x(x-8)+4x-(20)=0

We add all the numbers together, and all the variables

4x+2x(x-8)-20=0

We multiply parentheses

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

We add all the numbers together, and all the variables

2x^2-12x-20=0

a = 2; b = -12; c = -20;
Δ = b2-4ac
Δ = -122-4·2·(-20)
Δ = 304
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:

$\sqrt{\Delta}=\sqrt{304}=\sqrt{16*19}=\sqrt{16}*\sqrt{19}=4\sqrt{19}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{19}}{2*2}=\frac{12-4\sqrt{19}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{19}}{2*2}=\frac{12+4\sqrt{19}}{4}
$