A+b^2=256

Solution for A+b^2=256 equation:

+A^2=256

We move all terms to the left:

+A^2-(256)=0

a = 1; b = 0; c = -256;
Δ = b2-4ac
Δ = 02-4·1·(-256)
Δ = 1024
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1024}=32$
$A_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32}{2*1}=\frac{-32}{2}
=-16
$
$A_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32}{2*1}=\frac{32}{2}
=16
$