What are the solutions to the equation 4+5p-p^2=0 ?

The solutions to the equation 4+5p-p^2=0 are p=-(-5+sqrt(41))/2 ,p=(5+sqrt(41))/2

Find the zeros of 4+5p-p^2=0

The zeros of 4+5p-p^2=0 are p=-(-5+sqrt(41))/2 ,p=(5+sqrt(41))/2

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4 + 5 p - p^{2} = 0

p = - \frac{- 5 + 41}{2}, p = \frac{5 + 41}{2}

Decimal Notation

p = - 0.70156 \dots, p = 5.70156 \dots

Solution steps

4 + 5 p - p^{2} = 0

Write in the standard form

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

- p^{2} + 5 p + 4 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 ( - 1 ) \cdot 4}{2 ( - 1 )}

5^{2} - 4 (- 1) \cdot 4 = 41

p_{1, 2} = \frac{- 5 \pm 41}{2 ( - 1 )}

Separate the solutions

p_{1} = \frac{- 5 + 41}{2 ( - 1 )}, p_{2} = \frac{- 5 - 41}{2 ( - 1 )}

p = \frac{- 5 + 41}{2 ( - 1 )} : - \frac{- 5 + 41}{2}

p = \frac{- 5 - 41}{2 ( - 1 )} : \frac{5 + 41}{2}

The solutions to the quadratic equation are:

p = - \frac{- 5 + 41}{2}, p = \frac{5 + 41}{2}

\frac{x}{- 11} + 8 = 6

- 2 d^{2} + 16 d + 66

\frac{x}{x ^{2} - 1} < 0

30 - 53

e^{- \frac{1}{2} x}

What are the solutions to the equation 4+5p-p^2=0 ?
The solutions to the equation 4+5p-p^2=0 are p=-(-5+sqrt(41))/2 ,p=(5+sqrt(41))/2
Find the zeros of 4+5p-p^2=0
The zeros of 4+5p-p^2=0 are p=-(-5+sqrt(41))/2 ,p=(5+sqrt(41))/2