256=x^2+23x

Solution for 256=x^2+23x equation:

256=x^2+23x

We move all terms to the left:

256-(x^2+23x)=0

We get rid of parentheses

-x^2-23x+256=0

We add all the numbers together, and all the variables

-1x^2-23x+256=0

a = -1; b = -23; c = +256;
Δ = b2-4ac
Δ = -232-4·(-1)·256
Δ = 1553
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-23)-\sqrt{1553}}{2*-1}=\frac{23-\sqrt{1553}}{-2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-23)+\sqrt{1553}}{2*-1}=\frac{23+\sqrt{1553}}{-2}
$