2y+3y^2-47=0

Solution for 2y+3y^2-47=0 equation:

2y+3y^2-47=0

a = 3; b = 2; c = -47;
Δ = b2-4ac
Δ = 22-4·3·(-47)
Δ = 568
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:

$\sqrt{\Delta}=\sqrt{568}=\sqrt{4*142}=\sqrt{4}*\sqrt{142}=2\sqrt{142}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{142}}{2*3}=\frac{-2-2\sqrt{142}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{142}}{2*3}=\frac{-2+2\sqrt{142}}{6}
$