3y^2+8y-6=0

Solution for 3y^2+8y-6=0 equation:

3y^2+8y-6=0

a = 3; b = 8; c = -6;
Δ = b2-4ac
Δ = 82-4·3·(-6)
Δ = 136
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:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-2\sqrt{34}}{2*3}=\frac{-8-2\sqrt{34}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+2\sqrt{34}}{2*3}=\frac{-8+2\sqrt{34}}{6}
$