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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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


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

(+6Y^2+15Y)

We get rid of parentheses

6Y^2+15Y

Back to the equation:

-(6Y^2+15Y)


We get rid of parentheses

-6Y^2-15Y=0

a = -6; b = -15; c = 0;
Δ = b2-4ac
Δ = -152-4·(-6)·0
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-15}{2*-6}=\frac{0}{-12}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+15}{2*-6}=\frac{30}{-12}
=-2+1/2
$