f+f2+f=20

Solution for f+f2+f=20 equation:

f+f2+f=20

We move all terms to the left:

f+f2+f-(20)=0

We add all the numbers together, and all the variables

f^2+2f-20=0

a = 1; b = 2; c = -20;
Δ = b2-4ac
Δ = 22-4·1·(-20)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{21}}{2*1}=\frac{-2-2\sqrt{21}}{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{21}}{2*1}=\frac{-2+2\sqrt{21}}{2}
$