What are the solutions to the equation p^2+12p-55=-5 ?

The solutions to the equation p^2+12p-55=-5 are p=-6+sqrt(86),p=-6-sqrt(86)

Find the zeros of p^2+12p-55=-5

The zeros of p^2+12p-55=-5 are p=-6+sqrt(86),p=-6-sqrt(86)

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p^{2} + 12 p - 55 = - 5

p = - 6 + 86, p = - 6 - 86

Solution steps

p^{2} + 12 p - 55 = - 5

Move

5

to the left side

p^{2} + 12 p - 50 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 50 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 50) = 286

p_{1, 2} = \frac{- 12 \pm 2 86}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 12 + 2 86}{2 \cdot 1}, p_{2} = \frac{- 12 - 2 86}{2 \cdot 1}

p = \frac{- 12 + 2 86}{2 \cdot 1} : - 6 + 86

p = \frac{- 12 - 2 86}{2 \cdot 1} : - 6 - 86

The solutions to the quadratic equation are:

p = - 6 + 86, p = - 6 - 86

∣ x - 6 ∣ + 1 = 8

y^{2} + 3 = 0

4 t - 32 + t^{2}

x^{2} - 9 x + 18

(9 + 5 i) (9 - 5 i)

What are the solutions to the equation p^2+12p-55=-5 ?
The solutions to the equation p^2+12p-55=-5 are p=-6+sqrt(86),p=-6-sqrt(86)
Find the zeros of p^2+12p-55=-5
The zeros of p^2+12p-55=-5 are p=-6+sqrt(86),p=-6-sqrt(86)