y-5=3-y(y+2)

Solution for y-5=3-y(y+2) equation:

y-5=3-y(y+2)

We move all terms to the left:

y-5-(3-y(y+2))=0


We calculate terms in parentheses: -(3-y(y+2)), so:

3-y(y+2)

determiningTheFunctionDomain
-y(y+2)+3

We multiply parentheses

-y^2-2y+3

We add all the numbers together, and all the variables

-1y^2-2y+3

Back to the equation:

-(-1y^2-2y+3)


We get rid of parentheses

1y^2+2y+y-3-5=0

We add all the numbers together, and all the variables

y^2+3y-8=0

a = 1; b = 3; c = -8;
Δ = b2-4ac
Δ = 32-4·1·(-8)
Δ = 41
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-\sqrt{41}}{2*1}=\frac{-3-\sqrt{41}}{2}
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{41}}{2*1}=\frac{-3+\sqrt{41}}{2}
$