3r^2-9r-6=0

Solution for 3r^2-9r-6=0 equation:

3r^2-9r-6=0

a = 3; b = -9; c = -6;
Δ = b2-4ac
Δ = -92-4·3·(-6)
Δ = 153
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:

$\sqrt{\Delta}=\sqrt{153}=\sqrt{9*17}=\sqrt{9}*\sqrt{17}=3\sqrt{17}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-3\sqrt{17}}{2*3}=\frac{9-3\sqrt{17}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+3\sqrt{17}}{2*3}=\frac{9+3\sqrt{17}}{6}
$