3y(3+y)=5

Solution for 3y(3+y)=5 equation:

3y(3+y)=5

We move all terms to the left:

3y(3+y)-(5)=0

We add all the numbers together, and all the variables

3y(y+3)-5=0

We multiply parentheses

3y^2+9y-5=0

a = 3; b = 9; c = -5;
Δ = b2-4ac
Δ = 92-4·3·(-5)
Δ = 141
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{141}}{2*3}=\frac{-9-\sqrt{141}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{141}}{2*3}=\frac{-9+\sqrt{141}}{6}
$