2(y-6)=3(y-4)y

Solution for 2(y-6)=3(y-4)y equation:

2(y-6)=3(y-4)y

We move all terms to the left:

2(y-6)-(3(y-4)y)=0

We multiply parentheses

2y-(3(y-4)y)-12=0


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

3(y-4)y

We multiply parentheses

3y^2-12y

Back to the equation:

-(3y^2-12y)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-3y^2+14y-12=0

a = -3; b = 14; c = -12;
Δ = b2-4ac
Δ = 142-4·(-3)·(-12)
Δ = 52
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:

$\sqrt{\Delta}=\sqrt{52}=\sqrt{4*13}=\sqrt{4}*\sqrt{13}=2\sqrt{13}$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(14)-2\sqrt{13}}{2*-3}=\frac{-14-2\sqrt{13}}{-6}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+2\sqrt{13}}{2*-3}=\frac{-14+2\sqrt{13}}{-6}
$