f(9f-4)=0

Solution for f(9f-4)=0 equation:

f(9f-4)=0

We multiply parentheses

9f^2-4f=0

a = 9; b = -4; c = 0;
Δ = b2-4ac
Δ = -42-4·9·0
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4}{2*9}=\frac{0}{18}
=0
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4}{2*9}=\frac{8}{18}
=4/9
$