3r^2+3r-9=0

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

3r^2+3r-9=0

a = 3; b = 3; c = -9;
Δ = b2-4ac
Δ = 32-4·3·(-9)
Δ = 117
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:

$\sqrt{\Delta}=\sqrt{117}=\sqrt{9*13}=\sqrt{9}*\sqrt{13}=3\sqrt{13}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3\sqrt{13}}{2*3}=\frac{-3-3\sqrt{13}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3\sqrt{13}}{2*3}=\frac{-3+3\sqrt{13}}{6}
$