10.5^2+b^2=28^2

Solution for 10.5^2+b^2=28^2 equation:

10.5^2+b^2=28^2

We move all terms to the left:

10.5^2+b^2-(28^2)=0

We add all the numbers together, and all the variables

b^2-673.75=0

a = 1; b = 0; c = -673.75;
Δ = b2-4ac
Δ = 02-4·1·(-673.75)
Δ = 2695
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:

$\sqrt{\Delta}=\sqrt{2695}=\sqrt{49*55}=\sqrt{49}*\sqrt{55}=7\sqrt{55}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-7\sqrt{55}}{2*1}=\frac{0-7\sqrt{55}}{2}
=-\frac{7\sqrt{55}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+7\sqrt{55}}{2*1}=\frac{0+7\sqrt{55}}{2}
=\frac{7\sqrt{55}}{2}
$