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

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

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

We move all terms to the left:

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


We calculate terms in parentheses: -(2x+x(2x-4)), so:

2x+x(2x-4)

We multiply parentheses

2x^2+2x-4x

We add all the numbers together, and all the variables

2x^2-2x

Back to the equation:

-(2x^2-2x)


We get rid of parentheses

x^2-2x^2+2x=0

We add all the numbers together, and all the variables

-1x^2+2x=0

a = -1; b = 2; c = 0;
Δ = b2-4ac
Δ = 22-4·(-1)·0
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2}{2*-1}=\frac{-4}{-2}
=+2
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2}{2*-1}=\frac{0}{-2}
=0
$