What are the solutions to the equation p^2+12p-54=0 ?

The solutions to the equation p^2+12p-54=0 are p=3(sqrt(10)-2),p=-3(2+sqrt(10))

Find the zeros of p^2+12p-54=0

The zeros of p^2+12p-54=0 are p=3(sqrt(10)-2),p=-3(2+sqrt(10))

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p^{2} + 12 p - 54 = 0

p = 3 (10 - 2), p = - 3 (2 + 10)

Decimal Notation

p = 3.48683 \dots, p = - 15.48683 \dots

Solution steps

p^{2} + 12 p - 54 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 1 \cdot ( - 54 )}{2 \cdot 1}

1 2^{2} - 4 \cdot 1 \cdot (- 54) = 610

p_{1, 2} = \frac{- 12 \pm 6 10}{2 \cdot 1}

Separate the solutions

p_{1} = \frac{- 12 + 6 10}{2 \cdot 1}, p_{2} = \frac{- 12 - 6 10}{2 \cdot 1}

p = \frac{- 12 + 6 10}{2 \cdot 1} : 3 (10 - 2)

p = \frac{- 12 - 6 10}{2 \cdot 1} : - 3 (2 + 10)

The solutions to the quadratic equation are:

p = 3 (10 - 2), p = - 3 (2 + 10)

- 2 (x - 5)^{2} = (2 x + 1)^{2} - 57

x^{2} + x - 2 = 0

- 16 t^{2} + 700 = 0

9 x^{2} + 6 x = 0

12 x^{2} - 192 x + 527 = 0

What are the solutions to the equation p^2+12p-54=0 ?
The solutions to the equation p^2+12p-54=0 are p=3(sqrt(10)-2),p=-3(2+sqrt(10))
Find the zeros of p^2+12p-54=0
The zeros of p^2+12p-54=0 are p=3(sqrt(10)-2),p=-3(2+sqrt(10))