17^2+x^2=28^2

Solution for 17^2+x^2=28^2 equation:

17^2+x^2=28^2

We move all terms to the left:

17^2+x^2-(28^2)=0

We add all the numbers together, and all the variables

x^2-495=0

a = 1; b = 0; c = -495;
Δ = b2-4ac
Δ = 02-4·1·(-495)
Δ = 1980
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:

$\sqrt{\Delta}=\sqrt{1980}=\sqrt{36*55}=\sqrt{36}*\sqrt{55}=6\sqrt{55}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{55}}{2*1}=\frac{0-6\sqrt{55}}{2}
=-\frac{6\sqrt{55}}{2}
=-3\sqrt{55}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{55}}{2*1}=\frac{0+6\sqrt{55}}{2}
=\frac{6\sqrt{55}}{2}
=3\sqrt{55}
$