y(3y+4)=15

Solution for y(3y+4)=15 equation:

y(3y+4)=15

We move all terms to the left:

y(3y+4)-(15)=0

We multiply parentheses

3y^2+4y-15=0

a = 3; b = 4; c = -15;
Δ = b2-4ac
Δ = 42-4·3·(-15)
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-14}{2*3}=\frac{-18}{6}
=-3
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+14}{2*3}=\frac{10}{6}
=1+2/3
$