3y^2+2y-33=0

Solution for 3y^2+2y-33=0 equation:

3y^2+2y-33=0

a = 3; b = 2; c = -33;
Δ = b2-4ac
Δ = 22-4·3·(-33)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-20}{2*3}=\frac{-22}{6}
=-3+2/3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+20}{2*3}=\frac{18}{6}
=3
$