Y=(4x+2)(3x-4)

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+12Y^2-16Y+6Y-8))=0


We calculate terms in parentheses: -((+12Y^2-16Y+6Y-8)), so:

(+12Y^2-16Y+6Y-8)

We get rid of parentheses

12Y^2-16Y+6Y-8

We add all the numbers together, and all the variables

12Y^2-10Y-8

Back to the equation:

-(12Y^2-10Y-8)


We get rid of parentheses

-12Y^2+10Y+8=0

a = -12; b = 10; c = +8;
Δ = b2-4ac
Δ = 102-4·(-12)·8
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-22}{2*-12}=\frac{-32}{-24}
=1+1/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+22}{2*-12}=\frac{12}{-24}
=-1/2
$