x^2+25^2=32^2

Solution for x^2+25^2=32^2 equation:

x^2+25^2=32^2

We move all terms to the left:

x^2+25^2-(32^2)=0

We add all the numbers together, and all the variables

x^2-399=0

a = 1; b = 0; c = -399;
Δ = b2-4ac
Δ = 02-4·1·(-399)
Δ = 1596
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1596}=\sqrt{4*399}=\sqrt{4}*\sqrt{399}=2\sqrt{399}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{399}}{2*1}=\frac{0-2\sqrt{399}}{2}
=-\frac{2\sqrt{399}}{2}
=-\sqrt{399}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{399}}{2*1}=\frac{0+2\sqrt{399}}{2}
=\frac{2\sqrt{399}}{2}
=\sqrt{399}
$