3r2-4r-6=0

Solution for 3r2-4r-6=0 equation:

3r^2-4r-6=0

a = 3; b = -4; c = -6;
Δ = b2-4ac
Δ = -42-4·3·(-6)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-2\sqrt{22}}{2*3}=\frac{4-2\sqrt{22}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+2\sqrt{22}}{2*3}=\frac{4+2\sqrt{22}}{6}
$