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

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

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

We move all terms to the left:

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

We multiply parentheses ..

-((+2Y^2+10Y-2Y-10))=0


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

(+2Y^2+10Y-2Y-10)

We get rid of parentheses

2Y^2+10Y-2Y-10

We add all the numbers together, and all the variables

2Y^2+8Y-10

Back to the equation:

-(2Y^2+8Y-10)


We get rid of parentheses

-2Y^2-8Y+10=0

a = -2; b = -8; c = +10;
Δ = b2-4ac
Δ = -82-4·(-2)·10
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-12}{2*-2}=\frac{-4}{-4}
=1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+12}{2*-2}=\frac{20}{-4}
=-5
$