h^2+(h-7)(h+7)=13^2

Solution for h^2+(h-7)(h+7)=13^2 equation:

h^2+(h-7)(h+7)=13^2

We move all terms to the left:

h^2+(h-7)(h+7)-(13^2)=0

We add all the numbers together, and all the variables

h^2+(h-7)(h+7)-169=0

We use the square of the difference formula

h^2+h^2-49-169=0

We add all the numbers together, and all the variables

2h^2-218=0

a = 2; b = 0; c = -218;
Δ = b2-4ac
Δ = 02-4·2·(-218)
Δ = 1744
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:

$\sqrt{\Delta}=\sqrt{1744}=\sqrt{16*109}=\sqrt{16}*\sqrt{109}=4\sqrt{109}$
$h_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{109}}{2*2}=\frac{0-4\sqrt{109}}{4}
=-\frac{4\sqrt{109}}{4}
=-\sqrt{109}
$
$h_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{109}}{2*2}=\frac{0+4\sqrt{109}}{4}
=\frac{4\sqrt{109}}{4}
=\sqrt{109}
$