Y=(3x+1)(5x-4)

Solution for Y=(3x+1)(5x-4) equation:

=(3Y+1)(5Y-4)

We move all terms to the left:

-((3Y+1)(5Y-4))=0

We multiply parentheses ..

-((+15Y^2-12Y+5Y-4))=0


We calculate terms in parentheses: -((+15Y^2-12Y+5Y-4)), so:

(+15Y^2-12Y+5Y-4)

We get rid of parentheses

15Y^2-12Y+5Y-4

We add all the numbers together, and all the variables

15Y^2-7Y-4

Back to the equation:

-(15Y^2-7Y-4)


We get rid of parentheses

-15Y^2+7Y+4=0

a = -15; b = 7; c = +4;
Δ = b2-4ac
Δ = 72-4·(-15)·4
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-17}{2*-15}=\frac{-24}{-30}
=4/5
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+17}{2*-15}=\frac{10}{-30}
=-1/3
$