What are the solutions to the equation p^2+2p-13=-5 ?

The solutions to the equation p^2+2p-13=-5 are p=2,p=-4

Find the zeros of p^2+2p-13=-5

The zeros of p^2+2p-13=-5 are p=2,p=-4

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p^{2} + 2 p - 13 = - 5

p = 2, p = - 4

Solution steps

p^{2} + 2 p - 13 = - 5

Move

5

to the left side

p^{2} + 2 p - 8 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot ( - 8 )}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot (- 8) = 6

p_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 2 + 6}{2 \cdot 1}, p_{2} = \frac{- 2 - 6}{2 \cdot 1}

p = \frac{- 2 + 6}{2 \cdot 1} : 2

p = \frac{- 2 - 6}{2 \cdot 1} : - 4

The solutions to the quadratic equation are:

p = 2, p = - 4

528

3 x^{2} + 30 x + 75

a^{3} - 8 b^{3}

\frac{x ^{2}}{x ^{5}}

\frac{9 !}{5 !}

What are the solutions to the equation p^2+2p-13=-5 ?
The solutions to the equation p^2+2p-13=-5 are p=2,p=-4
Find the zeros of p^2+2p-13=-5
The zeros of p^2+2p-13=-5 are p=2,p=-4