What are the solutions to the equation p^2-2p-15=0 ?

The solutions to the equation p^2-2p-15=0 are p=5,p=-3

Find the zeros of p^2-2p-15=0

The zeros of p^2-2p-15=0 are p=5,p=-3

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p^{2} - 2 p - 15 = 0

p = 5, p = - 3

Solution steps

p^{2} - 2 p - 15 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 1 \cdot ( - 15 )}{2 \cdot 1}

(- 2)^{2} - 4 \cdot 1 \cdot (- 15) = 8

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

Separate the solutions

p_{1} = \frac{- ( - 2 ) + 8}{2 \cdot 1}, p_{2} = \frac{- ( - 2 ) - 8}{2 \cdot 1}

p = \frac{- ( - 2 ) + 8}{2 \cdot 1} : 5

p = \frac{- ( - 2 ) - 8}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

p = 5, p = - 3

- (\frac{3}{2})^{- 1}

827

\frac{x - 8}{x + 1} \geq 0

\frac{d}{2} + 1 > - 1

(\frac{4 f ^{3} g}{3 h ^{6}})^{3}

What are the solutions to the equation p^2-2p-15=0 ?
The solutions to the equation p^2-2p-15=0 are p=5,p=-3
Find the zeros of p^2-2p-15=0
The zeros of p^2-2p-15=0 are p=5,p=-3