Y=(3x+8)(4x-1)=

Solution for Y=(3x+8)(4x-1)= equation:

=(3Y+8)(4Y-1)=

We move all terms to the left:

-((3Y+8)(4Y-1))=0

We multiply parentheses ..

-((+12Y^2-3Y+32Y-8))=0


We calculate terms in parentheses: -((+12Y^2-3Y+32Y-8)), so:

(+12Y^2-3Y+32Y-8)

We get rid of parentheses

12Y^2-3Y+32Y-8

We add all the numbers together, and all the variables

12Y^2+29Y-8

Back to the equation:

-(12Y^2+29Y-8)


We get rid of parentheses

-12Y^2-29Y+8=0

a = -12; b = -29; c = +8;
Δ = b2-4ac
Δ = -292-4·(-12)·8
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-35}{2*-12}=\frac{-6}{-24}
=1/4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+35}{2*-12}=\frac{64}{-24}
=-2+2/3
$