x(200-4x)=160+20x

Solution for x(200-4x)=160+20x equation:

x(200-4x)=160+20x

We move all terms to the left:

x(200-4x)-(160+20x)=0

We add all the numbers together, and all the variables

x(-4x+200)-(20x+160)=0

We multiply parentheses

-4x^2+200x-(20x+160)=0

We get rid of parentheses

-4x^2+200x-20x-160=0

We add all the numbers together, and all the variables

-4x^2+180x-160=0

a = -4; b = 180; c = -160;
Δ = b2-4ac
Δ = 1802-4·(-4)·(-160)
Δ = 29840
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:

$\sqrt{\Delta}=\sqrt{29840}=\sqrt{16*1865}=\sqrt{16}*\sqrt{1865}=4\sqrt{1865}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(180)-4\sqrt{1865}}{2*-4}=\frac{-180-4\sqrt{1865}}{-8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(180)+4\sqrt{1865}}{2*-4}=\frac{-180+4\sqrt{1865}}{-8}
$