What are the solutions to the equation solvefor p,0=p+cp(1-p)-ep(1-p) ?

The solutions to the equation solvefor p,0=p+cp(1-p)-ep(1-p) are p=0,p=-(1+c-e)/(-c+e);\quad c\ne e

Find the zeros of solvefor p,0=p+cp(1-p)-ep(1-p)

The zeros of solvefor p,0=p+cp(1-p)-ep(1-p) are p=0,p=-(1+c-e)/(-c+e);\quad c\ne e

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solve for

p, 0 = p + c p (1 - p) - e p (1 - p)

p = 0, p = - \frac{1 + c - e}{- c + e}; c \neq = e

Solution steps

0 = p + c p (1 - p) - e p (1 - p)

Expand

p + c p (1 - p) - e p (1 - p) : p + c p - c p^{2} - e p + e p^{2}

0 = p + c p - c p^{2} - e p + e p^{2}

Switch sides

p + c p - c p^{2} - e p + e p^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

(- c + e) p^{2} + (1 + c - e) p = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( 1 + c - e ) \pm ( 1 + c - e ) ^{2} - 4 ( - c + e ) \cdot 0}{2 ( - c + e )}

Simplify

(1 + c - e)^{2} - 4 (- c + e) \cdot 0 : c + 1 - e

p_{1, 2} = \frac{- ( 1 + c - e ) \pm ( c + 1 - e )}{2 ( - c + e )}; c \neq = e

Separate the solutions

p_{1} = \frac{- ( 1 + c - e ) + c + 1 - e}{2 ( - c + e )}, p_{2} = \frac{- ( 1 + c - e ) - ( c + 1 - e )}{2 ( - c + e )}

p = \frac{- ( 1 + c - e ) + c + 1 - e}{2 ( - c + e )} : 0

p = \frac{- ( 1 + c - e ) - ( c + 1 - e )}{2 ( - c + e )} : - \frac{1 + c - e}{- c + e}

The solutions to the quadratic equation are:

p = 0, p = - \frac{1 + c - e}{- c + e}; c \neq = e

15 = - x^{2} + 8 x

5 x^{2} - 6 x + 7 = 0

x, y^{3} = x^{2} - 1

a^{2} + 4 a + 8 = 0

x^{2} - 6 x + 8 = 9

What are the solutions to the equation solvefor p,0=p+cp(1-p)-ep(1-p) ?
The solutions to the equation solvefor p,0=p+cp(1-p)-ep(1-p) are p=0,p=-(1+c-e)/(-c+e);\quad c\ne e
Find the zeros of solvefor p,0=p+cp(1-p)-ep(1-p)
The zeros of solvefor p,0=p+cp(1-p)-ep(1-p) are p=0,p=-(1+c-e)/(-c+e);\quad c\ne e