2x(x-5)=x^2-4x

Solution for 2x(x-5)=x^2-4x equation:

2x(x-5)=x^2-4x

We move all terms to the left:

2x(x-5)-(x^2-4x)=0

We multiply parentheses

2x^2-10x-(x^2-4x)=0

We get rid of parentheses

2x^2-x^2-10x+4x=0

We add all the numbers together, and all the variables

x^2-6x=0

a = 1; b = -6; c = 0;
Δ = b2-4ac
Δ = -62-4·1·0
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-6}{2*1}=\frac{0}{2}
=0
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+6}{2*1}=\frac{12}{2}
=6
$