What are the solutions to the equation p^2+8p+12=0 ?

The solutions to the equation p^2+8p+12=0 are p=-2,p=-6

Find the zeros of p^2+8p+12=0

The zeros of p^2+8p+12=0 are p=-2,p=-6

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p^{2} + 8 p + 12 = 0

p = - 2, p = - 6

Solution steps

p^{2} + 8 p + 12 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 1 \cdot 12}{2 \cdot 1}

8^{2} - 4 \cdot 1 \cdot 12 = 4

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

Separate the solutions

p_{1} = \frac{- 8 + 4}{2 \cdot 1}, p_{2} = \frac{- 8 - 4}{2 \cdot 1}

p = \frac{- 8 + 4}{2 \cdot 1} : - 2

p = \frac{- 8 - 4}{2 \cdot 1} : - 6

The solutions to the quadratic equation are:

p = - 2, p = - 6

20 x^{2} + 7 x - 3 = 0

12 x^{2} + x - 6 = 0

9 (2 m - 3)^{2} + 8 = 449

(3 + 2 i) (3 - 2 i)

\frac{x + 1}{2} = \frac{x}{7}

What are the solutions to the equation p^2+8p+12=0 ?
The solutions to the equation p^2+8p+12=0 are p=-2,p=-6
Find the zeros of p^2+8p+12=0
The zeros of p^2+8p+12=0 are p=-2,p=-6