3y2+2y-3=0

Solution for 3y2+2y-3=0 equation:

3y^2+2y-3=0

a = 3; b = 2; c = -3;
Δ = b2-4ac
Δ = 22-4·3·(-3)
Δ = 40
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{10}}{2*3}=\frac{-2-2\sqrt{10}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{10}}{2*3}=\frac{-2+2\sqrt{10}}{6}
$