9f^2-20f+4=0

Solution for 9f^2-20f+4=0 equation:

9f^2-20f+4=0

a = 9; b = -20; c = +4;
Δ = b2-4ac
Δ = -202-4·9·4
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$f_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-16}{2*9}=\frac{4}{18}
=2/9
$
$f_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+16}{2*9}=\frac{36}{18}
=2
$