(8+f)(8-f)=36

Solution for (8+f)(8-f)=36 equation:

(8+f)(8-f)=36

We move all terms to the left:

(8+f)(8-f)-(36)=0

We add all the numbers together, and all the variables

(f+8)(-1f+8)-36=0

We multiply parentheses ..

(-1f^2+8f-8f+64)-36=0

We get rid of parentheses

-1f^2+8f-8f+64-36=0

We add all the numbers together, and all the variables

-1f^2+28=0

a = -1; b = 0; c = +28;
Δ = b2-4ac
Δ = 02-4·(-1)·28
Δ = 112
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{112}=\sqrt{16*7}=\sqrt{16}*\sqrt{7}=4\sqrt{7}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{7}}{2*-1}=\frac{0-4\sqrt{7}}{-2}
=-\frac{4\sqrt{7}}{-2}
=-\frac{2\sqrt{7}}{-1}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{7}}{2*-1}=\frac{0+4\sqrt{7}}{-2}
=\frac{4\sqrt{7}}{-2}
=\frac{2\sqrt{7}}{-1}
$