-3y(y-1)+2y=4(y-3)

Solution for -3y(y-1)+2y=4(y-3) equation:

-3y(y-1)+2y=4(y-3)

We move all terms to the left:

-3y(y-1)+2y-(4(y-3))=0

We add all the numbers together, and all the variables

2y-3y(y-1)-(4(y-3))=0

We multiply parentheses

-3y^2+2y+3y-(4(y-3))=0


We calculate terms in parentheses: -(4(y-3)), so:

4(y-3)

We multiply parentheses

4y-12

Back to the equation:

-(4y-12)


We add all the numbers together, and all the variables

-3y^2+5y-(4y-12)=0

We get rid of parentheses

-3y^2+5y-4y+12=0

We add all the numbers together, and all the variables

-3y^2+y+12=0

a = -3; b = 1; c = +12;
Δ = b2-4ac
Δ = 12-4·(-3)·12
Δ = 145
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{-(1)-\sqrt{145}}{2*-3}=\frac{-1-\sqrt{145}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+\sqrt{145}}{2*-3}=\frac{-1+\sqrt{145}}{-6}
$