What are the solutions to the equation -4p^2+12000p=0 ?

The solutions to the equation -4p^2+12000p=0 are p=0,p=3000

Find the zeros of -4p^2+12000p=0

The zeros of -4p^2+12000p=0 are p=0,p=3000

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- 4 p^{2} + 12000 p = 0

p = 0, p = 3000

Solution steps

- 4 p^{2} + 12000 p = 0

Solve with the quadratic formula

p_{1, 2} = \frac{- 12000 \pm 1200 0 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

1200 0^{2} - 4 (- 4) \cdot 0 = 12000

p_{1, 2} = \frac{- 12000 \pm 12000}{2 ( - 4 )}

Separate the solutions

p_{1} = \frac{- 12000 + 12000}{2 ( - 4 )}, p_{2} = \frac{- 12000 - 12000}{2 ( - 4 )}

p = \frac{- 12000 + 12000}{2 ( - 4 )} : 0

p = \frac{- 12000 - 12000}{2 ( - 4 )} : 3000

The solutions to the quadratic equation are:

p = 0, p = 3000

2 - 3 b > \frac{11 - 15 b}{7}

5 x + 11 \geq 1 + 15 x

x^{2} - 10 x = 100

- 4 (x + 3)^{2} = 28

6 (- 3 v + 1) = 5 (- 2 v - 2)

What are the solutions to the equation -4p^2+12000p=0 ?
The solutions to the equation -4p^2+12000p=0 are p=0,p=3000
Find the zeros of -4p^2+12000p=0
The zeros of -4p^2+12000p=0 are p=0,p=3000