h^2+25=21^2

Solution for h^2+25=21^2 equation:

h^2+25=21^2

We move all terms to the left:

h^2+25-(21^2)=0

We add all the numbers together, and all the variables

h^2-416=0

a = 1; b = 0; c = -416;
Δ = b2-4ac
Δ = 02-4·1·(-416)
Δ = 1664
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1664}=\sqrt{64*26}=\sqrt{64}*\sqrt{26}=8\sqrt{26}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{26}}{2*1}=\frac{0-8\sqrt{26}}{2}
=-\frac{8\sqrt{26}}{2}
=-4\sqrt{26}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{26}}{2*1}=\frac{0+8\sqrt{26}}{2}
=\frac{8\sqrt{26}}{2}
=4\sqrt{26}
$