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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6Y^2+4Y+3Y+2))=0


We calculate terms in parentheses: -((+6Y^2+4Y+3Y+2)), so:

(+6Y^2+4Y+3Y+2)

We get rid of parentheses

6Y^2+4Y+3Y+2

We add all the numbers together, and all the variables

6Y^2+7Y+2

Back to the equation:

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


We get rid of parentheses

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

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

$\sqrt{\Delta}=\sqrt{1}=1$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-1}{2*-6}=\frac{6}{-12}
=-1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+1}{2*-6}=\frac{8}{-12}
=-2/3
$