25+b^2=256

Solution for 25+b^2=256 equation:

25+b^2=256

We move all terms to the left:

25+b^2-(256)=0

We add all the numbers together, and all the variables

b^2-231=0

a = 1; b = 0; c = -231;
Δ = b2-4ac
Δ = 02-4·1·(-231)
Δ = 924
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{924}=\sqrt{4*231}=\sqrt{4}*\sqrt{231}=2\sqrt{231}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{231}}{2*1}=\frac{0-2\sqrt{231}}{2}
=-\frac{2\sqrt{231}}{2}
=-\sqrt{231}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{231}}{2*1}=\frac{0+2\sqrt{231}}{2}
=\frac{2\sqrt{231}}{2}
=\sqrt{231}
$