3^2+b^2=25^2

Solution for 3^2+b^2=25^2 equation:

3^2+b^2=25^2

We move all terms to the left:

3^2+b^2-(25^2)=0

We add all the numbers together, and all the variables

b^2-616=0

a = 1; b = 0; c = -616;
Δ = b2-4ac
Δ = 02-4·1·(-616)
Δ = 2464
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:

$\sqrt{\Delta}=\sqrt{2464}=\sqrt{16*154}=\sqrt{16}*\sqrt{154}=4\sqrt{154}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{154}}{2*1}=\frac{0-4\sqrt{154}}{2}
=-\frac{4\sqrt{154}}{2}
=-2\sqrt{154}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{154}}{2*1}=\frac{0+4\sqrt{154}}{2}
=\frac{4\sqrt{154}}{2}
=2\sqrt{154}
$