What are the solutions to the equation -2p^2-42p=32 ?

The solutions to the equation -2p^2-42p=32 are p=-(21+sqrt(377))/2 ,p=-(21-sqrt(377))/2

Find the zeros of -2p^2-42p=32

The zeros of -2p^2-42p=32 are p=-(21+sqrt(377))/2 ,p=-(21-sqrt(377))/2

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- 2 p^{2} - 42 p = 32

p = - \frac{21 + 377}{2}, p = - \frac{21 - 377}{2}

Decimal Notation

p = - 20.20824 \dots, p = - 0.79175 \dots

Solution steps

- 2 p^{2} - 42 p = 32

Move

32

to the left side

- 2 p^{2} - 42 p - 32 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 42 ) \pm ( - 42 ) ^{2} - 4 ( - 2 ) ( - 32 )}{2 ( - 2 )}

(- 42)^{2} - 4 (- 2) (- 32) = 2377

p_{1, 2} = \frac{- ( - 42 ) \pm 2 377}{2 ( - 2 )}

Separate the solutions

p_{1} = \frac{- ( - 42 ) + 2 377}{2 ( - 2 )}, p_{2} = \frac{- ( - 42 ) - 2 377}{2 ( - 2 )}

p = \frac{- ( - 42 ) + 2 377}{2 ( - 2 )} : - \frac{21 + 377}{2}

p = \frac{- ( - 42 ) - 2 377}{2 ( - 2 )} : - \frac{21 - 377}{2}

The solutions to the quadratic equation are:

p = - \frac{21 + 377}{2}, p = - \frac{21 - 377}{2}

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

a^{2} + 8^{2} = 9^{2}

- 12 x^{2} + 96 x + 100 = 0

b (x) = - 5 x^{2}

4 = 3 x^{2} + 14 x + 4

What are the solutions to the equation -2p^2-42p=32 ?
The solutions to the equation -2p^2-42p=32 are p=-(21+sqrt(377))/2 ,p=-(21-sqrt(377))/2
Find the zeros of -2p^2-42p=32
The zeros of -2p^2-42p=32 are p=-(21+sqrt(377))/2 ,p=-(21-sqrt(377))/2