0=(2-y)(8-y)

Solution for 0=(2-y)(8-y) equation:

0=(2-y)(8-y)

We move all terms to the left:

0-((2-y)(8-y))=0

We add all the numbers together, and all the variables

-((-1y+2)(-1y+8))+0=0

We add all the numbers together, and all the variables

-((-1y+2)(-1y+8))=0

We multiply parentheses ..

-((+y^2-8y-2y+16))=0


We calculate terms in parentheses: -((+y^2-8y-2y+16)), so:

(+y^2-8y-2y+16)

We get rid of parentheses

y^2-8y-2y+16

We add all the numbers together, and all the variables

y^2-10y+16

Back to the equation:

-(y^2-10y+16)


We get rid of parentheses

-y^2+10y-16=0

We add all the numbers together, and all the variables

-1y^2+10y-16=0

a = -1; b = 10; c = -16;
Δ = b2-4ac
Δ = 102-4·(-1)·(-16)
Δ = 36
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{36}=6$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(10)-6}{2*-1}=\frac{-16}{-2}
=+8
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+6}{2*-1}=\frac{-4}{-2}
=+2
$