What are the solutions to the equation 12p^2+8=7 ?

The solutions to the equation 12p^2+8=7 are p= 1/(2sqrt(3))i,p=-1/(2sqrt(3))i

Find the zeros of 12p^2+8=7

The zeros of 12p^2+8=7 are p= 1/(2sqrt(3))i,p=-1/(2sqrt(3))i

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12 p^{2} + 8 = 7

p = \frac{1}{2 3} i, p = - \frac{1}{2 3} i

Solution steps

12 p^{2} + 8 = 7

Move

8

to the right side

12 p^{2} = - 1

Divide both sides by

12

p^{2} = - \frac{1}{12}

For

x^{2} = f (a)

the solutions are

x = f (a), - f (a)

p = - \frac{1}{12}, p = - - \frac{1}{12}

- \frac{1}{12} = \frac{1}{2 3} i

- - \frac{1}{12} = - \frac{1}{2 3} i

p = \frac{1}{2 3} i, p = - \frac{1}{2 3} i

e^{x} (e^{x})

- 3 x^{2} + 7 x + 6 = 0

3 x^{\frac{3}{2}} - 9 x^{\frac{1}{2}} + 6 x^{- \frac{1}{2}}

- 8 \leq 4 x - 2 < 0

\frac{x ^{2} - 3 x - 4}{x ^{2} + 6 x + 5} < 0

What are the solutions to the equation 12p^2+8=7 ?
The solutions to the equation 12p^2+8=7 are p= 1/(2sqrt(3))i,p=-1/(2sqrt(3))i
Find the zeros of 12p^2+8=7
The zeros of 12p^2+8=7 are p= 1/(2sqrt(3))i,p=-1/(2sqrt(3))i