3y(y+1)+3=5

Solution for 3y(y+1)+3=5 equation:

3y(y+1)+3=5

We move all terms to the left:

3y(y+1)+3-(5)=0

We add all the numbers together, and all the variables

3y(y+1)-2=0

We multiply parentheses

3y^2+3y-2=0

a = 3; b = 3; c = -2;
Δ = b2-4ac
Δ = 32-4·3·(-2)
Δ = 33
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{-(3)-\sqrt{33}}{2*3}=\frac{-3-\sqrt{33}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{33}}{2*3}=\frac{-3+\sqrt{33}}{6}
$