f(2)-F(6)=20

Solution for f(2)-F(6)=20 equation:

f(2)-(6)=20

We move all terms to the left:

f(2)-(6)-(20)=0

We add all the numbers together, and all the variables

f^2-26=0

a = 1; b = 0; c = -26;
Δ = b2-4ac
Δ = 02-4·1·(-26)
Δ = 104
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:

$\sqrt{\Delta}=\sqrt{104}=\sqrt{4*26}=\sqrt{4}*\sqrt{26}=2\sqrt{26}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{26}}{2*1}=\frac{0-2\sqrt{26}}{2}
=-\frac{2\sqrt{26}}{2}
=-\sqrt{26}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{26}}{2*1}=\frac{0+2\sqrt{26}}{2}
=\frac{2\sqrt{26}}{2}
=\sqrt{26}
$