6r^2+2r-3=0

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

6r^2+2r-3=0

a = 6; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·6·(-3)
Δ = 76
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:

$\sqrt{\Delta}=\sqrt{76}=\sqrt{4*19}=\sqrt{4}*\sqrt{19}=2\sqrt{19}$
$r_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{19}}{2*6}=\frac{-2-2\sqrt{19}}{12}
$
$r_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{19}}{2*6}=\frac{-2+2\sqrt{19}}{12}
$