x^2-2x+12=2x(x+10)

Solution for x^2-2x+12=2x(x+10) equation:

x^2-2x+12=2x(x+10)

We move all terms to the left:

x^2-2x+12-(2x(x+10))=0


We calculate terms in parentheses: -(2x(x+10)), so:

2x(x+10)

We multiply parentheses

2x^2+20x

Back to the equation:

-(2x^2+20x)


We get rid of parentheses

x^2-2x^2-2x-20x+12=0

We add all the numbers together, and all the variables

-1x^2-22x+12=0

a = -1; b = -22; c = +12;
Δ = b2-4ac
Δ = -222-4·(-1)·12
Δ = 532
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:

$\sqrt{\Delta}=\sqrt{532}=\sqrt{4*133}=\sqrt{4}*\sqrt{133}=2\sqrt{133}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-2\sqrt{133}}{2*-1}=\frac{22-2\sqrt{133}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+2\sqrt{133}}{2*-1}=\frac{22+2\sqrt{133}}{-2}
$