6r2+6r-15=0

Solution for 6r2+6r-15=0 equation:

6r^2+6r-15=0

a = 6; b = 6; c = -15;
Δ = b2-4ac
Δ = 62-4·6·(-15)
Δ = 396
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:

$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{11}}{2*6}=\frac{-6-6\sqrt{11}}{12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{11}}{2*6}=\frac{-6+6\sqrt{11}}{12}
$