3(y-2)+15=-3y(y-3)+6y

Solution for 3(y-2)+15=-3y(y-3)+6y equation:

3(y-2)+15=-3y(y-3)+6y

We move all terms to the left:

3(y-2)+15-(-3y(y-3)+6y)=0

We multiply parentheses

3y-(-3y(y-3)+6y)-6+15=0


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

-3y(y-3)+6y

We add all the numbers together, and all the variables

6y-3y(y-3)

We multiply parentheses

-3y^2+6y+9y

We add all the numbers together, and all the variables

-3y^2+15y

Back to the equation:

-(-3y^2+15y)


We add all the numbers together, and all the variables

-(-3y^2+15y)+3y+9=0

We get rid of parentheses

3y^2-15y+3y+9=0

We add all the numbers together, and all the variables

3y^2-12y+9=0

a = 3; b = -12; c = +9;
Δ = b2-4ac
Δ = -122-4·3·9
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6}{2*3}=\frac{6}{6}
=1
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6}{2*3}=\frac{18}{6}
=3
$