a^2+25^2=38^2

Solution for a^2+25^2=38^2 equation:

a^2+25^2=38^2

We move all terms to the left:

a^2+25^2-(38^2)=0

We add all the numbers together, and all the variables

a^2-819=0

a = 1; b = 0; c = -819;
Δ = b2-4ac
Δ = 02-4·1·(-819)
Δ = 3276
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:

$\sqrt{\Delta}=\sqrt{3276}=\sqrt{36*91}=\sqrt{36}*\sqrt{91}=6\sqrt{91}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{91}}{2*1}=\frac{0-6\sqrt{91}}{2}
=-\frac{6\sqrt{91}}{2}
=-3\sqrt{91}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{91}}{2*1}=\frac{0+6\sqrt{91}}{2}
=\frac{6\sqrt{91}}{2}
=3\sqrt{91}
$