What are the solutions to the equation 2p^2+24=-16p ?

The solutions to the equation 2p^2+24=-16p are p=-2,p=-6

Find the zeros of 2p^2+24=-16p

The zeros of 2p^2+24=-16p are p=-2,p=-6

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2 p^{2} + 24 = - 16 p

p = - 2, p = - 6

Solution steps

2 p^{2} + 24 = - 16 p

Move

16 p

to the left side

2 p^{2} + 24 + 16 p = 0

Write in the standard form

a x^{2} + b x + c = 0

2 p^{2} + 16 p + 24 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 \cdot 2 \cdot 24}{2 \cdot 2}

1 6^{2} - 4 \cdot 2 \cdot 24 = 8

p_{1, 2} = \frac{- 16 \pm 8}{2 \cdot 2}

Separate the solutions

p_{1} = \frac{- 16 + 8}{2 \cdot 2}, p_{2} = \frac{- 16 - 8}{2 \cdot 2}

p = \frac{- 16 + 8}{2 \cdot 2} : - 2

p = \frac{- 16 - 8}{2 \cdot 2} : - 6

The solutions to the quadratic equation are:

p = - 2, p = - 6

- x^{4} + 11 x^{2} + 18 x + 8

\frac{5 x - 3}{2} > 6

0. 9^{5}

\frac{6 - 11 i}{12 - 5 i}

7 y + 8 = - 2 y - 64

What are the solutions to the equation 2p^2+24=-16p ?
The solutions to the equation 2p^2+24=-16p are p=-2,p=-6
Find the zeros of 2p^2+24=-16p
The zeros of 2p^2+24=-16p are p=-2,p=-6