12^2+b^2=23^2

Solution for 12^2+b^2=23^2 equation:

12^2+b^2=23^2

We move all terms to the left:

12^2+b^2-(23^2)=0

We add all the numbers together, and all the variables

b^2-385=0

a = 1; b = 0; c = -385;
Δ = b2-4ac
Δ = 02-4·1·(-385)
Δ = 1540
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1540}=\sqrt{4*385}=\sqrt{4}*\sqrt{385}=2\sqrt{385}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{385}}{2*1}=\frac{0-2\sqrt{385}}{2}
=-\frac{2\sqrt{385}}{2}
=-\sqrt{385}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{385}}{2*1}=\frac{0+2\sqrt{385}}{2}
=\frac{2\sqrt{385}}{2}
=\sqrt{385}
$