20x=(8x-4)(5x+5)

Solution for 20x=(8x-4)(5x+5) equation:

20x=(8x-4)(5x+5)

We move all terms to the left:

20x-((8x-4)(5x+5))=0

We multiply parentheses ..

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


We calculate terms in parentheses: -((+40x^2+40x-20x-20)), so:

(+40x^2+40x-20x-20)

We get rid of parentheses

40x^2+40x-20x-20

We add all the numbers together, and all the variables

40x^2+20x-20

Back to the equation:

-(40x^2+20x-20)


We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-40x^2+20=0

a = -40; b = 0; c = +20;
Δ = b2-4ac
Δ = 02-4·(-40)·20
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{-(0)-40\sqrt{2}}{2*-40}=\frac{0-40\sqrt{2}}{-80}
=-\frac{40\sqrt{2}}{-80}
=-\frac{\sqrt{2}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40\sqrt{2}}{2*-40}=\frac{0+40\sqrt{2}}{-80}
=\frac{40\sqrt{2}}{-80}
=\frac{\sqrt{2}}{-2}
$