x+14=x^2-32x+256

Solution for x+14=x^2-32x+256 equation:

x+14=x^2-32x+256

We move all terms to the left:

x+14-(x^2-32x+256)=0

We get rid of parentheses

-x^2+x+32x-256+14=0

We add all the numbers together, and all the variables

-1x^2+33x-242=0

a = -1; b = 33; c = -242;
Δ = b2-4ac
Δ = 332-4·(-1)·(-242)
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(33)-11}{2*-1}=\frac{-44}{-2}
=+22
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(33)+11}{2*-1}=\frac{-22}{-2}
=+11
$