5x^2-40x=5x(8-x)

Solution for 5x^2-40x=5x(8-x) equation:

5x^2-40x=5x(8-x)

We move all terms to the left:

5x^2-40x-(5x(8-x))=0

We add all the numbers together, and all the variables

5x^2-40x-(5x(-1x+8))=0


We calculate terms in parentheses: -(5x(-1x+8)), so:

5x(-1x+8)

We multiply parentheses

-5x^2+40x

Back to the equation:

-(-5x^2+40x)


We get rid of parentheses

5x^2+5x^2-40x-40x=0

We add all the numbers together, and all the variables

10x^2-80x=0

a = 10; b = -80; c = 0;
Δ = b2-4ac
Δ = -802-4·10·0
Δ = 6400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{6400}=80$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-80}{2*10}=\frac{0}{20}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+80}{2*10}=\frac{160}{20}
=8
$