What are the solutions to the equation p^2+12p=0 ?

The solutions to the equation p^2+12p=0 are p=0,p=-12

Find the zeros of p^2+12p=0

The zeros of p^2+12p=0 are p=0,p=-12

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p^{2} + 12 p = 0

p = 0, p = - 12

Solution steps

p^{2} + 12 p = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 0}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 0 = 12

p_{1, 2} = \frac{- 12 \pm 12}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 12 + 12}{2 \cdot 1}, p_{2} = \frac{- 12 - 12}{2 \cdot 1}

p = \frac{- 12 + 12}{2 \cdot 1} : 0

p = \frac{- 12 - 12}{2 \cdot 1} : - 12

The solutions to the quadratic equation are:

p = 0, p = - 12

(7)^{2} - 7 (7) + 6 = 0

6 \leq x \leq 8

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30 z^{2} + 13 z - 3

63 x^{2} y - 7 y

What are the solutions to the equation p^2+12p=0 ?
The solutions to the equation p^2+12p=0 are p=0,p=-12
Find the zeros of p^2+12p=0
The zeros of p^2+12p=0 are p=0,p=-12