What are the solutions to the equation -3x^2+12x=0 ?

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

Find the zeros of -3x^2+12x=0

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

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

x = 0, x = 4

Solution steps

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

1 2^{2} - 4 (- 3) \cdot 0 = 12

x_{1, 2} = \frac{- 12 \pm 12}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 12 + 12}{2 ( - 3 )}, x_{2} = \frac{- 12 - 12}{2 ( - 3 )}

x = \frac{- 12 + 12}{2 ( - 3 )} : 0

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

The solutions to the quadratic equation are:

x = 0, x = 4

- 2 x^{2} + 10 x - 3 = 0

x^{2} + 16 x = - 55

6 x^{2} - 12 x + 1 = 0

x^{2} + 7 = 88

x^{2} + 10 = 7 x

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