15=2y+3y^2

Solution for 15=2y+3y^2 equation:

15=2y+3y^2

We move all terms to the left:

15-(2y+3y^2)=0

We get rid of parentheses

-3y^2-2y+15=0

a = -3; b = -2; c = +15;
Δ = b2-4ac
Δ = -22-4·(-3)·15
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{46}}{2*-3}=\frac{2-2\sqrt{46}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{46}}{2*-3}=\frac{2+2\sqrt{46}}{-6}
$