14^2+b^2=23^2

Solution for 14^2+b^2=23^2 equation:

14^2+b^2=23^2

We move all terms to the left:

14^2+b^2-(23^2)=0

We add all the numbers together, and all the variables

b^2-333=0

a = 1; b = 0; c = -333;
Δ = b2-4ac
Δ = 02-4·1·(-333)
Δ = 1332
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:

$\sqrt{\Delta}=\sqrt{1332}=\sqrt{36*37}=\sqrt{36}*\sqrt{37}=6\sqrt{37}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{37}}{2*1}=\frac{0-6\sqrt{37}}{2}
=-\frac{6\sqrt{37}}{2}
=-3\sqrt{37}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{37}}{2*1}=\frac{0+6\sqrt{37}}{2}
=\frac{6\sqrt{37}}{2}
=3\sqrt{37}
$