5r^2-18r+9=0

Solution for 5r^2-18r+9=0 equation:

5r^2-18r+9=0

a = 5; b = -18; c = +9;
Δ = b2-4ac
Δ = -182-4·5·9
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-12}{2*5}=\frac{6}{10}
=3/5
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+12}{2*5}=\frac{30}{10}
=3
$