23^2+10^2=x^2

Solution for 23^2+10^2=x^2 equation:

23^2+10^2=x^2

We move all terms to the left:

23^2+10^2-(x^2)=0

We add all the numbers together, and all the variables

-1x^2+629=0

a = -1; b = 0; c = +629;
Δ = b2-4ac
Δ = 02-4·(-1)·629
Δ = 2516
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:

$\sqrt{\Delta}=\sqrt{2516}=\sqrt{4*629}=\sqrt{4}*\sqrt{629}=2\sqrt{629}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{629}}{2*-1}=\frac{0-2\sqrt{629}}{-2}
=-\frac{2\sqrt{629}}{-2}
=-\frac{\sqrt{629}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{629}}{2*-1}=\frac{0+2\sqrt{629}}{-2}
=\frac{2\sqrt{629}}{-2}
=\frac{\sqrt{629}}{-1}
$