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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+6Y^2+10Y-21Y-35))=0


We calculate terms in parentheses: -((+6Y^2+10Y-21Y-35)), so:

(+6Y^2+10Y-21Y-35)

We get rid of parentheses

6Y^2+10Y-21Y-35

We add all the numbers together, and all the variables

6Y^2-11Y-35

Back to the equation:

-(6Y^2-11Y-35)


We get rid of parentheses

-6Y^2+11Y+35=0

a = -6; b = 11; c = +35;
Δ = b2-4ac
Δ = 112-4·(-6)·35
Δ = 961
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{961}=31$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-31}{2*-6}=\frac{-42}{-12}
=3+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+31}{2*-6}=\frac{20}{-12}
=-1+2/3
$