What are the solutions to the equation -p^2+2p+3=0 ?

The solutions to the equation -p^2+2p+3=0 are p=-1,p=3

Find the zeros of -p^2+2p+3=0

The zeros of -p^2+2p+3=0 are p=-1,p=3

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- p^{2} + 2 p + 3 = 0

p = - 1, p = 3

Solution steps

- p^{2} + 2 p + 3 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 3 = 4

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

Separate the solutions

p_{1} = \frac{- 2 + 4}{2 ( - 1 )}, p_{2} = \frac{- 2 - 4}{2 ( - 1 )}

p = \frac{- 2 + 4}{2 ( - 1 )} : - 1

p = \frac{- 2 - 4}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

p = - 1, p = 3

15 + 2 (2.5 - x) = 11 + x^{2}

x, (x - 2)^{2} + (y - 3)^{2} = 0

(3 m - 32) \cdot x^{2} - (m - 4) x + 4 = 0

9 = 4 + x^{2}

5 x (x - 3) = 2

What are the solutions to the equation -p^2+2p+3=0 ?
The solutions to the equation -p^2+2p+3=0 are p=-1,p=3
Find the zeros of -p^2+2p+3=0
The zeros of -p^2+2p+3=0 are p=-1,p=3