What are the solutions to the equation p^2+15p+56=0 ?

The solutions to the equation p^2+15p+56=0 are p=-7,p=-8

Find the zeros of p^2+15p+56=0

The zeros of p^2+15p+56=0 are p=-7,p=-8

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p^{2} + 15 p + 56 = 0

p = - 7, p = - 8

Solution steps

p^{2} + 15 p + 56 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 15 \pm 1 5 ^{2} - 4 \cdot 1 \cdot 56}{2 \cdot 1}

1 5^{2} - 4 \cdot 1 \cdot 56 = 1

p_{1, 2} = \frac{- 15 \pm 1}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 15 + 1}{2 \cdot 1}, p_{2} = \frac{- 15 - 1}{2 \cdot 1}

p = \frac{- 15 + 1}{2 \cdot 1} : - 7

p = \frac{- 15 - 1}{2 \cdot 1} : - 8

The solutions to the quadratic equation are:

p = - 7, p = - 8

x^{2} - 0.5 = 0

6 x^{2} - 37 x - 35 = 0

4 x^{2} - 6 x + 10 = 0

a, a^{2} - 2 ab + b^{2} = 1.40885

x^{2} - 3 x = 2 x - 6

What are the solutions to the equation p^2+15p+56=0 ?
The solutions to the equation p^2+15p+56=0 are p=-7,p=-8
Find the zeros of p^2+15p+56=0
The zeros of p^2+15p+56=0 are p=-7,p=-8