2x^=x^2+x

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

2x^=x^2+x

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

-x^2+2x-x=0

We add all the numbers together, and all the variables

-1x^2+x=0

a = -1; b = 1; c = 0;
Δ = b2-4ac
Δ = 12-4·(-1)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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