f^2+20f=41

Solution for f^2+20f=41 equation:

f^2+20f=41

We move all terms to the left:

f^2+20f-(41)=0

a = 1; b = 20; c = -41;
Δ = b2-4ac
Δ = 202-4·1·(-41)
Δ = 564
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:

$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-2\sqrt{141}}{2*1}=\frac{-20-2\sqrt{141}}{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+2\sqrt{141}}{2*1}=\frac{-20+2\sqrt{141}}{2}
$