12(y+2)=6(y2+4)

Solution for 12(y+2)=6(y2+4) equation:

12(y+2)=6(y2+4)

We move all terms to the left:

12(y+2)-(6(y2+4))=0

We add all the numbers together, and all the variables

-(6(+y^2+4))+12(y+2)=0

We multiply parentheses

-(6(+y^2+4))+12y+24=0


We calculate terms in parentheses: -(6(+y^2+4)), so:

6(+y^2+4)

We multiply parentheses

6y^2+24

Back to the equation:

-(6y^2+24)


We add all the numbers together, and all the variables

12y-(6y^2+24)+24=0

We get rid of parentheses

-6y^2+12y-24+24=0

We add all the numbers together, and all the variables

-6y^2+12y=0

a = -6; b = 12; c = 0;
Δ = b2-4ac
Δ = 122-4·(-6)·0
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-12}{2*-6}=\frac{-24}{-12}
=+2
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+12}{2*-6}=\frac{0}{-12}
=0
$