Y=(3x-5)(2x-4)

Solution for Y=(3x-5)(2x-4) equation:

=(3Y-5)(2Y-4)

We move all terms to the left:

-((3Y-5)(2Y-4))=0

We multiply parentheses ..

-((+6Y^2-12Y-10Y+20))=0


We calculate terms in parentheses: -((+6Y^2-12Y-10Y+20)), so:

(+6Y^2-12Y-10Y+20)

We get rid of parentheses

6Y^2-12Y-10Y+20

We add all the numbers together, and all the variables

6Y^2-22Y+20

Back to the equation:

-(6Y^2-22Y+20)


We get rid of parentheses

-6Y^2+22Y-20=0

a = -6; b = 22; c = -20;
Δ = b2-4ac
Δ = 222-4·(-6)·(-20)
Δ = 4
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4}=2$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(22)-2}{2*-6}=\frac{-24}{-12}
=+2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(22)+2}{2*-6}=\frac{-20}{-12}
=1+2/3
$