2y=y(3-2y)+6

Solution for 2y=y(3-2y)+6 equation:

2y=y(3-2y)+6

We move all terms to the left:

2y-(y(3-2y)+6)=0

We add all the numbers together, and all the variables

2y-(y(-2y+3)+6)=0


We calculate terms in parentheses: -(y(-2y+3)+6), so:

y(-2y+3)+6

We multiply parentheses

-2y^2+3y+6

Back to the equation:

-(-2y^2+3y+6)


We get rid of parentheses

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

We add all the numbers together, and all the variables

2y^2-1y-6=0

a = 2; b = -1; c = -6;
Δ = b2-4ac
Δ = -12-4·2·(-6)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*2}=\frac{-6}{4}
=-1+1/2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*2}=\frac{8}{4}
=2
$