28^2+22^2=x^2

Solution for 28^2+22^2=x^2 equation:

28^2+22^2=x^2

We move all terms to the left:

28^2+22^2-(x^2)=0

We add all the numbers together, and all the variables

-1x^2+1268=0

a = -1; b = 0; c = +1268;
Δ = b2-4ac
Δ = 02-4·(-1)·1268
Δ = 5072
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:

$\sqrt{\Delta}=\sqrt{5072}=\sqrt{16*317}=\sqrt{16}*\sqrt{317}=4\sqrt{317}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{317}}{2*-1}=\frac{0-4\sqrt{317}}{-2}
=-\frac{4\sqrt{317}}{-2}
=-\frac{2\sqrt{317}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{317}}{2*-1}=\frac{0+4\sqrt{317}}{-2}
=\frac{4\sqrt{317}}{-2}
=\frac{2\sqrt{317}}{-1}
$