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

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

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

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

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

p = 0, p = 3000

Solution steps

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

Switch sides

- 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

3 y^{2} - 6 y - 45 = 0

6 x^{2} - 30 x + 24

(- \frac{4}{5})^{- 4}

x^{7} - 1

5 x + 4 = 19

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