Y=(3x-2)(x+8)

Solution for Y=(3x-2)(x+8) equation:

=(3Y-2)(Y+8)

We move all terms to the left:

-((3Y-2)(Y+8))=0

We multiply parentheses ..

-((+3Y^2+24Y-2Y-16))=0


We calculate terms in parentheses: -((+3Y^2+24Y-2Y-16)), so:

(+3Y^2+24Y-2Y-16)

We get rid of parentheses

3Y^2+24Y-2Y-16

We add all the numbers together, and all the variables

3Y^2+22Y-16

Back to the equation:

-(3Y^2+22Y-16)


We get rid of parentheses

-3Y^2-22Y+16=0

a = -3; b = -22; c = +16;
Δ = b2-4ac
Δ = -222-4·(-3)·16
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-22)-26}{2*-3}=\frac{-4}{-6}
=2/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-22)+26}{2*-3}=\frac{48}{-6}
=-8
$