Y=X(1+p)+p^2

Solution for Y=X(1+p)+p^2 equation:

=Y(1+Y)+Y^2

We move all terms to the left:

-(Y(1+Y)+Y^2)=0

We add all the numbers together, and all the variables

-(Y(Y+1)+Y^2)=0


We calculate terms in parentheses: -(Y(Y+1)+Y^2), so:

Y(Y+1)+Y^2

determiningTheFunctionDomain
Y^2+Y(Y+1)

We multiply parentheses

Y^2+Y^2+Y

We add all the numbers together, and all the variables

2Y^2+Y

Back to the equation:

-(2Y^2+Y)


We get rid of parentheses

-2Y^2-Y=0

We add all the numbers together, and all the variables

-2Y^2-1Y=0

a = -2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-2)·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-2}=\frac{0}{-4}
=0
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-2}=\frac{2}{-4}
=-1/2
$