14^2+b^2=28^2

Solution for 14^2+b^2=28^2 equation:

14^2+b^2=28^2

We move all terms to the left:

14^2+b^2-(28^2)=0

We add all the numbers together, and all the variables

b^2-588=0

a = 1; b = 0; c = -588;
Δ = b2-4ac
Δ = 02-4·1·(-588)
Δ = 2352
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:

$\sqrt{\Delta}=\sqrt{2352}=\sqrt{784*3}=\sqrt{784}*\sqrt{3}=28\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-28\sqrt{3}}{2*1}=\frac{0-28\sqrt{3}}{2}
=-\frac{28\sqrt{3}}{2}
=-14\sqrt{3}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+28\sqrt{3}}{2*1}=\frac{0+28\sqrt{3}}{2}
=\frac{28\sqrt{3}}{2}
=14\sqrt{3}
$