Y=(2x-5)(4x+7)

Solution for Y=(2x-5)(4x+7) equation:

=(2Y-5)(4Y+7)

We move all terms to the left:

-((2Y-5)(4Y+7))=0

We multiply parentheses ..

-((+8Y^2+14Y-20Y-35))=0


We calculate terms in parentheses: -((+8Y^2+14Y-20Y-35)), so:

(+8Y^2+14Y-20Y-35)

We get rid of parentheses

8Y^2+14Y-20Y-35

We add all the numbers together, and all the variables

8Y^2-6Y-35

Back to the equation:

-(8Y^2-6Y-35)


We get rid of parentheses

-8Y^2+6Y+35=0

a = -8; b = 6; c = +35;
Δ = b2-4ac
Δ = 62-4·(-8)·35
Δ = 1156
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1156}=34$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-34}{2*-8}=\frac{-40}{-16}
=2+1/2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+34}{2*-8}=\frac{28}{-16}
=-1+3/4
$