3r^2+6r-4=0

Solution for 3r^2+6r-4=0 equation:

3r^2+6r-4=0

a = 3; b = 6; c = -4;
Δ = b2-4ac
Δ = 62-4·3·(-4)
Δ = 84
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:

$\sqrt{\Delta}=\sqrt{84}=\sqrt{4*21}=\sqrt{4}*\sqrt{21}=2\sqrt{21}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{21}}{2*3}=\frac{-6-2\sqrt{21}}{6}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{21}}{2*3}=\frac{-6+2\sqrt{21}}{6}
$