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

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

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

We move all terms to the left:

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


Domain of the equation: Y))!=0

Y!=0/1

Y!=0

Y∈R


We add all the numbers together, and all the variables

-((5Y-3)(+4Y-3/Y))=0

We multiply parentheses ..

-((+20Y^2-15Y^2-12Y+9Y))=0


We calculate terms in parentheses: -((+20Y^2-15Y^2-12Y+9Y)), so:

(+20Y^2-15Y^2-12Y+9Y)

We get rid of parentheses

20Y^2-15Y^2-12Y+9Y

We add all the numbers together, and all the variables

5Y^2-3Y

Back to the equation:

-(5Y^2-3Y)


We get rid of parentheses

-5Y^2+3Y=0

a = -5; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-5)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*-5}=\frac{-6}{-10}
=3/5
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*-5}=\frac{0}{-10}
=0
$