(3x-20)(100-x)=100

Solution for (3x-20)(100-x)=100 equation:

(3x-20)(100-x)=100

We move all terms to the left:

(3x-20)(100-x)-(100)=0

We add all the numbers together, and all the variables

(3x-20)(-1x+100)-100=0

We multiply parentheses ..

(-3x^2+300x+20x-2000)-100=0

We get rid of parentheses

-3x^2+300x+20x-2000-100=0

We add all the numbers together, and all the variables

-3x^2+320x-2100=0

a = -3; b = 320; c = -2100;
Δ = b2-4ac
Δ = 3202-4·(-3)·(-2100)
Δ = 77200
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:

$\sqrt{\Delta}=\sqrt{77200}=\sqrt{400*193}=\sqrt{400}*\sqrt{193}=20\sqrt{193}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(320)-20\sqrt{193}}{2*-3}=\frac{-320-20\sqrt{193}}{-6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(320)+20\sqrt{193}}{2*-3}=\frac{-320+20\sqrt{193}}{-6}
$