21^2=17^2+x^2

Solution for 21^2=17^2+x^2 equation:

21^2=17^2+x^2

We move all terms to the left:

21^2-(17^2+x^2)=0

We add all the numbers together, and all the variables

-(17^2+x^2)+441=0

We get rid of parentheses

-x^2+441-17^2=0

We add all the numbers together, and all the variables

-1x^2+152=0

a = -1; b = 0; c = +152;
Δ = b2-4ac
Δ = 02-4·(-1)·152
Δ = 608
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:

$\sqrt{\Delta}=\sqrt{608}=\sqrt{16*38}=\sqrt{16}*\sqrt{38}=4\sqrt{38}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{38}}{2*-1}=\frac{0-4\sqrt{38}}{-2}
=-\frac{4\sqrt{38}}{-2}
=-\frac{2\sqrt{38}}{-1}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{38}}{2*-1}=\frac{0+4\sqrt{38}}{-2}
=\frac{4\sqrt{38}}{-2}
=\frac{2\sqrt{38}}{-1}
$