What are the solutions to the equation -10p^2-2p=5 ?

The solutions to the equation -10p^2-2p=5 are p=-1/10-i 7/10 ,p=-1/10+i 7/10

Find the zeros of -10p^2-2p=5

The zeros of -10p^2-2p=5 are p=-1/10-i 7/10 ,p=-1/10+i 7/10

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- 10 p^{2} - 2 p = 5

p = - \frac{1}{10} - i \frac{7}{10}, p = - \frac{1}{10} + i \frac{7}{10}

Solution steps

- 10 p^{2} - 2 p = 5

Move

5

to the left side

- 10 p^{2} - 2 p - 5 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 10 ) ( - 5 )}{2 ( - 10 )}

Simplify

(- 2)^{2} - 4 (- 10) (- 5) : 14 i

p_{1, 2} = \frac{- ( - 2 ) \pm 14 i}{2 ( - 10 )}

Separate the solutions

p_{1} = \frac{- ( - 2 ) + 14 i}{2 ( - 10 )}, p_{2} = \frac{- ( - 2 ) - 14 i}{2 ( - 10 )}

p = \frac{- ( - 2 ) + 14 i}{2 ( - 10 )} : - \frac{1}{10} - i \frac{7}{10}

p = \frac{- ( - 2 ) - 14 i}{2 ( - 10 )} : - \frac{1}{10} + i \frac{7}{10}

The solutions to the quadratic equation are:

p = - \frac{1}{10} - i \frac{7}{10}, p = - \frac{1}{10} + i \frac{7}{10}

\frac{x}{6} + 3 = 7

7 x^{5} - 21 x^{4} + 14 x^{3}

\frac{x}{5} - 2 = - 9

0 = x^{2} - 6 x - 4

5 (\frac{1}{3})

What are the solutions to the equation -10p^2-2p=5 ?
The solutions to the equation -10p^2-2p=5 are p=-1/10-i 7/10 ,p=-1/10+i 7/10
Find the zeros of -10p^2-2p=5
The zeros of -10p^2-2p=5 are p=-1/10-i 7/10 ,p=-1/10+i 7/10