a^2+15^2=21^2

Solution for a^2+15^2=21^2 equation:

a^2+15^2=21^2

We move all terms to the left:

a^2+15^2-(21^2)=0

We add all the numbers together, and all the variables

a^2-216=0

a = 1; b = 0; c = -216;
Δ = b2-4ac
Δ = 02-4·1·(-216)
Δ = 864
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{864}=\sqrt{144*6}=\sqrt{144}*\sqrt{6}=12\sqrt{6}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-12\sqrt{6}}{2*1}=\frac{0-12\sqrt{6}}{2}
=-\frac{12\sqrt{6}}{2}
=-6\sqrt{6}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+12\sqrt{6}}{2*1}=\frac{0+12\sqrt{6}}{2}
=\frac{12\sqrt{6}}{2}
=6\sqrt{6}
$