What are the solutions to the equation 12x-4x^2=0 ?

The solutions to the equation 12x-4x^2=0 are x=0,x=3

Find the zeros of 12x-4x^2=0

The zeros of 12x-4x^2=0 are x=0,x=3

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12 x - 4 x^{2} = 0

x = 0, x = 3

Solution steps

12 x - 4 x^{2} = 0

Write in the standard form

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

- 4 x^{2} + 12 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 4 ) \cdot 0}{2 ( - 4 )}

1 2^{2} - 4 (- 4) \cdot 0 = 12

x_{1, 2} = \frac{- 12 \pm 12}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 12 + 12}{2 ( - 4 )}, x_{2} = \frac{- 12 - 12}{2 ( - 4 )}

x = \frac{- 12 + 12}{2 ( - 4 )} : 0

x = \frac{- 12 - 12}{2 ( - 4 )} : 3

The solutions to the quadratic equation are:

x = 0, x = 3

- 4 + 11

6 v^{2} + 12 v = 0

x^{3} + 6 x^{2} - 24 x - 64

0.0172 x - 0.0476 = 0

\frac{2 π}{\frac{π}{8}}

What are the solutions to the equation 12x-4x^2=0 ?
The solutions to the equation 12x-4x^2=0 are x=0,x=3
Find the zeros of 12x-4x^2=0
The zeros of 12x-4x^2=0 are x=0,x=3