Y=(3x-5)(4x-6)

Solution for Y=(3x-5)(4x-6) equation:

=(3Y-5)(4Y-6)

We move all terms to the left:

-((3Y-5)(4Y-6))=0

We multiply parentheses ..

-((+12Y^2-18Y-20Y+30))=0


We calculate terms in parentheses: -((+12Y^2-18Y-20Y+30)), so:

(+12Y^2-18Y-20Y+30)

We get rid of parentheses

12Y^2-18Y-20Y+30

We add all the numbers together, and all the variables

12Y^2-38Y+30

Back to the equation:

-(12Y^2-38Y+30)


We get rid of parentheses

-12Y^2+38Y-30=0

a = -12; b = 38; c = -30;
Δ = b2-4ac
Δ = 382-4·(-12)·(-30)
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{-(38)-2}{2*-12}=\frac{-40}{-24}
=1+2/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(38)+2}{2*-12}=\frac{-36}{-24}
=1+1/2
$