Y=(3x+1)(2x-5)

Solution for Y=(3x+1)(2x-5) equation:

=(3Y+1)(2Y-5)

We move all terms to the left:

-((3Y+1)(2Y-5))=0

We multiply parentheses ..

-((+6Y^2-15Y+2Y-5))=0


We calculate terms in parentheses: -((+6Y^2-15Y+2Y-5)), so:

(+6Y^2-15Y+2Y-5)

We get rid of parentheses

6Y^2-15Y+2Y-5

We add all the numbers together, and all the variables

6Y^2-13Y-5

Back to the equation:

-(6Y^2-13Y-5)


We get rid of parentheses

-6Y^2+13Y+5=0

a = -6; b = 13; c = +5;
Δ = b2-4ac
Δ = 132-4·(-6)·5
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{-(13)-17}{2*-6}=\frac{-30}{-12}
=2+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(13)+17}{2*-6}=\frac{4}{-12}
=-1/3
$