What are the solutions to the equation p^2+12p=-5 ?

The solutions to the equation p^2+12p=-5 are p=-6+sqrt(31),p=-6-sqrt(31)

Find the zeros of p^2+12p=-5

The zeros of p^2+12p=-5 are p=-6+sqrt(31),p=-6-sqrt(31)

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p^{2} + 12 p = - 5

p = - 6 + 31, p = - 6 - 31

Solution steps

p^{2} + 12 p = - 5

Move

5

to the left side

p^{2} + 12 p + 5 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot 5 = 231

p_{1, 2} = \frac{- 12 \pm 2 31}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 12 + 2 31}{2 \cdot 1}, p_{2} = \frac{- 12 - 2 31}{2 \cdot 1}

p = \frac{- 12 + 2 31}{2 \cdot 1} : - 6 + 31

p = \frac{- 12 - 2 31}{2 \cdot 1} : - 6 - 31

The solutions to the quadratic equation are:

p = - 6 + 31, p = - 6 - 31

2304

n^{2} + 5 n + 6

\frac{y ^{7}}{y ^{4}}

uv \cdot 4 u v^{5}

3 x^{2} - 5 x < - 2

What are the solutions to the equation p^2+12p=-5 ?
The solutions to the equation p^2+12p=-5 are p=-6+sqrt(31),p=-6-sqrt(31)
Find the zeros of p^2+12p=-5
The zeros of p^2+12p=-5 are p=-6+sqrt(31),p=-6-sqrt(31)