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

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

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

We move all terms to the left:

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

We multiply parentheses

3y^2-9y-((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^2-9y-y-3=0

We add all the numbers together, and all the variables

3y^2-10y-3=0

a = 3; b = -10; c = -3;
Δ = b2-4ac
Δ = -102-4·3·(-3)
Δ = 136
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{136}=\sqrt{4*34}=\sqrt{4}*\sqrt{34}=2\sqrt{34}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{34}}{2*3}=\frac{10-2\sqrt{34}}{6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{34}}{2*3}=\frac{10+2\sqrt{34}}{6}
$