Y=(8+2x)(10+2x)

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

=(8+2Y)(10+2Y)

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

-((+4Y^2+20Y+16Y+80))=0


We calculate terms in parentheses: -((+4Y^2+20Y+16Y+80)), so:

(+4Y^2+20Y+16Y+80)

We get rid of parentheses

4Y^2+20Y+16Y+80

We add all the numbers together, and all the variables

4Y^2+36Y+80

Back to the equation:

-(4Y^2+36Y+80)


We get rid of parentheses

-4Y^2-36Y-80=0

a = -4; b = -36; c = -80;
Δ = b2-4ac
Δ = -362-4·(-4)·(-80)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{16}=4$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4}{2*-4}=\frac{32}{-8}
=-4
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4}{2*-4}=\frac{40}{-8}
=-5
$