3y^2+12y-9=0

Solution for 3y^2+12y-9=0 equation:

3y^2+12y-9=0

a = 3; b = 12; c = -9;
Δ = b2-4ac
Δ = 122-4·3·(-9)
Δ = 252
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:

$\sqrt{\Delta}=\sqrt{252}=\sqrt{36*7}=\sqrt{36}*\sqrt{7}=6\sqrt{7}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{7}}{2*3}=\frac{-12-6\sqrt{7}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{7}}{2*3}=\frac{-12+6\sqrt{7}}{6}
$