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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6Y^2-10Y+3Y-5))=0


We calculate terms in parentheses: -((+6Y^2-10Y+3Y-5)), so:

(+6Y^2-10Y+3Y-5)

We get rid of parentheses

6Y^2-10Y+3Y-5

We add all the numbers together, and all the variables

6Y^2-7Y-5

Back to the equation:

-(6Y^2-7Y-5)


We get rid of parentheses

-6Y^2+7Y+5=0

a = -6; b = 7; c = +5;
Δ = b2-4ac
Δ = 72-4·(-6)·5
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*-6}=\frac{-20}{-12}
=1+2/3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-6}=\frac{6}{-12}
=-1/2
$