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

The solutions to the equation -p^2+2p+4=0 are p=1-sqrt(5),p=1+sqrt(5)

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

The zeros of -p^2+2p+4=0 are p=1-sqrt(5),p=1+sqrt(5)

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

p = 1 - 5, p = 1 + 5

Decimal Notation

p = - 1.23606 \dots, p = 3.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 (- 1) \cdot 4 = 25

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

p = 1 - 5, p = 1 + 5

x^{2} + 24 = - 10 x

12 x - 6 x^{2} = 0

- x^{2} + 8 x + 15 = 0

5 x^{2} + 7 x - 9 = 0

- 3 t^{2} + 18 t - 15 = 0

What are the solutions to the equation -p^2+2p+4=0 ?
The solutions to the equation -p^2+2p+4=0 are p=1-sqrt(5),p=1+sqrt(5)
Find the zeros of -p^2+2p+4=0
The zeros of -p^2+2p+4=0 are p=1-sqrt(5),p=1+sqrt(5)