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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

-((2Y-3)(-1Y+4))=0

We multiply parentheses ..

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


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

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

-2Y^2+11Y-12

Back to the equation:

-(-2Y^2+11Y-12)


We get rid of parentheses

2Y^2-11Y+12=0

a = 2; b = -11; c = +12;
Δ = b2-4ac
Δ = -112-4·2·12
Δ = 25
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{25}=5$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-11)-5}{2*2}=\frac{6}{4}
=1+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-11)+5}{2*2}=\frac{16}{4}
=4
$