a^2+23^2=29^2

Solution for a^2+23^2=29^2 equation:

a^2+23^2=29^2

We move all terms to the left:

a^2+23^2-(29^2)=0

We add all the numbers together, and all the variables

a^2-312=0

a = 1; b = 0; c = -312;
Δ = b2-4ac
Δ = 02-4·1·(-312)
Δ = 1248
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:

$\sqrt{\Delta}=\sqrt{1248}=\sqrt{16*78}=\sqrt{16}*\sqrt{78}=4\sqrt{78}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{78}}{2*1}=\frac{0-4\sqrt{78}}{2}
=-\frac{4\sqrt{78}}{2}
=-2\sqrt{78}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{78}}{2*1}=\frac{0+4\sqrt{78}}{2}
=\frac{4\sqrt{78}}{2}
=2\sqrt{78}
$