Y=(3x+7)(2x-5)

Solution for Y=(3x+7)(2x-5) equation:

=(3Y+7)(2Y-5)

We move all terms to the left:

-((3Y+7)(2Y-5))=0

We multiply parentheses ..

-((+6Y^2-15Y+14Y-35))=0


We calculate terms in parentheses: -((+6Y^2-15Y+14Y-35)), so:

(+6Y^2-15Y+14Y-35)

We get rid of parentheses

6Y^2-15Y+14Y-35

We add all the numbers together, and all the variables

6Y^2-1Y-35

Back to the equation:

-(6Y^2-1Y-35)


We get rid of parentheses

-6Y^2+1Y+35=0

We add all the numbers together, and all the variables

-6Y^2+Y+35=0

a = -6; b = 1; c = +35;
Δ = b2-4ac
Δ = 12-4·(-6)·35
Δ = 841
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{841}=29$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-29}{2*-6}=\frac{-30}{-12}
=2+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+29}{2*-6}=\frac{28}{-12}
=-2+1/3
$