3r^2-20r+9=0

Solution for 3r^2-20r+9=0 equation:

3r^2-20r+9=0

a = 3; b = -20; c = +9;
Δ = b2-4ac
Δ = -202-4·3·9
Δ = 292
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{292}=\sqrt{4*73}=\sqrt{4}*\sqrt{73}=2\sqrt{73}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{73}}{2*3}=\frac{20-2\sqrt{73}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{73}}{2*3}=\frac{20+2\sqrt{73}}{6}
$