What are the solutions to the equation -5p^2-40p=64 ?

The solutions to the equation -5p^2-40p=64 are p=-(4(5+sqrt(5)))/5 ,p=-(4(5-sqrt(5)))/5

Find the zeros of -5p^2-40p=64

The zeros of -5p^2-40p=64 are p=-(4(5+sqrt(5)))/5 ,p=-(4(5-sqrt(5)))/5

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- 5 p^{2} - 40 p = 64

p = - \frac{4 ( 5 + 5 )}{5}, p = - \frac{4 ( 5 - 5 )}{5}

Decimal Notation

p = - 5.78885 \dots, p = - 2.21114 \dots

Solution steps

- 5 p^{2} - 40 p = 64

Move

64

to the left side

- 5 p^{2} - 40 p - 64 = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- ( - 40 ) \pm ( - 40 ) ^{2} - 4 ( - 5 ) ( - 64 )}{2 ( - 5 )}

(- 40)^{2} - 4 (- 5) (- 64) = 85

p_{1, 2} = \frac{- ( - 40 ) \pm 8 5}{2 ( - 5 )}

Separate the solutions

p_{1} = \frac{- ( - 40 ) + 8 5}{2 ( - 5 )}, p_{2} = \frac{- ( - 40 ) - 8 5}{2 ( - 5 )}

p = \frac{- ( - 40 ) + 8 5}{2 ( - 5 )} : - \frac{4 ( 5 + 5 )}{5}

p = \frac{- ( - 40 ) - 8 5}{2 ( - 5 )} : - \frac{4 ( 5 - 5 )}{5}

The solutions to the quadratic equation are:

p = - \frac{4 ( 5 + 5 )}{5}, p = - \frac{4 ( 5 - 5 )}{5}

4 n^{2} - 30 n = - 36

\frac{1}{2} \cdot 17 \cdot 5

∣ x - 6 ∣ < 4

x (3 x + 1) - 5 (1 - 2 x) = (3 x + 2)^{2} + 1

What are the solutions to the equation -5p^2-40p=64 ?
The solutions to the equation -5p^2-40p=64 are p=-(4(5+sqrt(5)))/5 ,p=-(4(5-sqrt(5)))/5
Find the zeros of -5p^2-40p=64
The zeros of -5p^2-40p=64 are p=-(4(5+sqrt(5)))/5 ,p=-(4(5-sqrt(5)))/5