Y=(2x-8)(5x-10)

Solution for Y=(2x-8)(5x-10) equation:

=(2Y-8)(5Y-10)

We move all terms to the left:

-((2Y-8)(5Y-10))=0

We multiply parentheses ..

-((+10Y^2-20Y-40Y+80))=0


We calculate terms in parentheses: -((+10Y^2-20Y-40Y+80)), so:

(+10Y^2-20Y-40Y+80)

We get rid of parentheses

10Y^2-20Y-40Y+80

We add all the numbers together, and all the variables

10Y^2-60Y+80

Back to the equation:

-(10Y^2-60Y+80)


We get rid of parentheses

-10Y^2+60Y-80=0

a = -10; b = 60; c = -80;
Δ = b2-4ac
Δ = 602-4·(-10)·(-80)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(60)-20}{2*-10}=\frac{-80}{-20}
=+4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(60)+20}{2*-10}=\frac{-40}{-20}
=+2
$