3(1+2y)y-y^2=8

Solution for 3(1+2y)y-y^2=8 equation:

3(1+2y)y-y^2=8

We move all terms to the left:

3(1+2y)y-y^2-(8)=0

We add all the numbers together, and all the variables

-y^2+3(2y+1)y-8=0

We add all the numbers together, and all the variables

-1y^2+3(2y+1)y-8=0

We multiply parentheses

-1y^2+6y^2+3y-8=0

We add all the numbers together, and all the variables

5y^2+3y-8=0

a = 5; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·5·(-8)
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-13}{2*5}=\frac{-16}{10}
=-1+3/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+13}{2*5}=\frac{10}{10}
=1
$