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

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

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

We move all terms to the left:

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

We multiply parentheses

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


We calculate terms in parentheses: -(y(y-2)), so:

y(y-2)

We multiply parentheses

y^2-2y

Back to the equation:

-(y^2-2y)


We get rid of parentheses

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

We add all the numbers together, and all the variables

-1y^2+5y+12=0

a = -1; b = 5; c = +12;
Δ = b2-4ac
Δ = 52-4·(-1)·12
Δ = 73
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{-(5)-\sqrt{73}}{2*-1}=\frac{-5-\sqrt{73}}{-2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*-1}=\frac{-5+\sqrt{73}}{-2}
$