2x+4x(20-x)=56

Solution for 2x+4x(20-x)=56 equation:

2x+4x(20-x)=56

We move all terms to the left:

2x+4x(20-x)-(56)=0

We add all the numbers together, and all the variables

2x+4x(-1x+20)-56=0

We multiply parentheses

-4x^2+2x+80x-56=0

We add all the numbers together, and all the variables

-4x^2+82x-56=0

a = -4; b = 82; c = -56;
Δ = b2-4ac
Δ = 822-4·(-4)·(-56)
Δ = 5828
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:

$\sqrt{\Delta}=\sqrt{5828}=\sqrt{4*1457}=\sqrt{4}*\sqrt{1457}=2\sqrt{1457}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(82)-2\sqrt{1457}}{2*-4}=\frac{-82-2\sqrt{1457}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(82)+2\sqrt{1457}}{2*-4}=\frac{-82+2\sqrt{1457}}{-8}
$