(4x+20)(x-5)=40x

Solution for (4x+20)(x-5)=40x equation:

(4x+20)(x-5)=40x

We move all terms to the left:

(4x+20)(x-5)-(40x)=0

We add all the numbers together, and all the variables

-40x+(4x+20)(x-5)=0

We multiply parentheses ..

(+4x^2-20x+20x-100)-40x=0

We get rid of parentheses

4x^2-20x+20x-40x-100=0

We add all the numbers together, and all the variables

4x^2-40x-100=0

a = 4; b = -40; c = -100;
Δ = b2-4ac
Δ = -402-4·4·(-100)
Δ = 3200
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:

$\sqrt{\Delta}=\sqrt{3200}=\sqrt{1600*2}=\sqrt{1600}*\sqrt{2}=40\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-40\sqrt{2}}{2*4}=\frac{40-40\sqrt{2}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+40\sqrt{2}}{2*4}=\frac{40+40\sqrt{2}}{8}
$