a^2+24^2=32^2

Solution for a^2+24^2=32^2 equation:

a^2+24^2=32^2

We move all terms to the left:

a^2+24^2-(32^2)=0

We add all the numbers together, and all the variables

a^2-448=0

a = 1; b = 0; c = -448;
Δ = b2-4ac
Δ = 02-4·1·(-448)
Δ = 1792
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:

$\sqrt{\Delta}=\sqrt{1792}=\sqrt{256*7}=\sqrt{256}*\sqrt{7}=16\sqrt{7}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16\sqrt{7}}{2*1}=\frac{0-16\sqrt{7}}{2}
=-\frac{16\sqrt{7}}{2}
=-8\sqrt{7}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16\sqrt{7}}{2*1}=\frac{0+16\sqrt{7}}{2}
=\frac{16\sqrt{7}}{2}
=8\sqrt{7}
$