x^2+20x=2x^2+10x

Solution for x^2+20x=2x^2+10x equation:

x^2+20x=2x^2+10x

We move all terms to the left:

x^2+20x-(2x^2+10x)=0

We get rid of parentheses

x^2-2x^2+20x-10x=0

We add all the numbers together, and all the variables

-1x^2+10x=0

a = -1; b = 10; c = 0;
Δ = b2-4ac
Δ = 102-4·(-1)·0
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-10}{2*-1}=\frac{-20}{-2}
=+10
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+10}{2*-1}=\frac{0}{-2}
=0
$