f(f-1)=40

Solution for f(f-1)=40 equation:

f(f-1)=40

We move all terms to the left:

f(f-1)-(40)=0

We multiply parentheses

f^2-1f-40=0

a = 1; b = -1; c = -40;
Δ = b2-4ac
Δ = -12-4·1·(-40)
Δ = 161
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{161}}{2*1}=\frac{1-\sqrt{161}}{2}
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{161}}{2*1}=\frac{1+\sqrt{161}}{2}
$