5^2+b^2=25^2

Solution for 5^2+b^2=25^2 equation:

5^2+b^2=25^2

We move all terms to the left:

5^2+b^2-(25^2)=0

We add all the numbers together, and all the variables

b^2-600=0

a = 1; b = 0; c = -600;
Δ = b2-4ac
Δ = 02-4·1·(-600)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*1}=\frac{0-20\sqrt{6}}{2}
=-\frac{20\sqrt{6}}{2}
=-10\sqrt{6}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*1}=\frac{0+20\sqrt{6}}{2}
=\frac{20\sqrt{6}}{2}
=10\sqrt{6}
$