Y=(2/3x+4)(6x+2)

Solution for Y=(2/3x+4)(6x+2) equation:

=(2/3Y+4)(6Y+2)

We move all terms to the left:

-((2/3Y+4)(6Y+2))=0


Domain of the equation: 3Y+4)(6Y+2))!=0

Y∈R


We multiply parentheses ..

-((+12Y^2+4Y+24Y+8))=0


We calculate terms in parentheses: -((+12Y^2+4Y+24Y+8)), so:

(+12Y^2+4Y+24Y+8)

We get rid of parentheses

12Y^2+4Y+24Y+8

We add all the numbers together, and all the variables

12Y^2+28Y+8

Back to the equation:

-(12Y^2+28Y+8)


We get rid of parentheses

-12Y^2-28Y-8=0

a = -12; b = -28; c = -8;
Δ = b2-4ac
Δ = -282-4·(-12)·(-8)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-28)-20}{2*-12}=\frac{8}{-24}
=-1/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-28)+20}{2*-12}=\frac{48}{-24}
=-2
$