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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

(y+3)

We get rid of parentheses

y+3

Back to the equation:

-(y+3)


We get rid of parentheses

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

We add all the numbers together, and all the variables

3y^2-7y-3=0

a = 3; b = -7; c = -3;
Δ = b2-4ac
Δ = -72-4·3·(-3)
Δ = 85
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{-(-7)-\sqrt{85}}{2*3}=\frac{7-\sqrt{85}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+\sqrt{85}}{2*3}=\frac{7+\sqrt{85}}{6}
$