What are the solutions to the equation -0.3x^2+12x=0 ?

The solutions to the equation -0.3x^2+12x=0 are x=0,x=40

Find the zeros of -0.3x^2+12x=0

The zeros of -0.3x^2+12x=0 are x=0,x=40

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

x = 0, x = 40

Solution steps

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

Multiply both sides by

10

- 3 x^{2} + 120 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 120 \pm 12 0 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

12 0^{2} - 4 (- 3) \cdot 0 = 120

x_{1, 2} = \frac{- 120 \pm 120}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 120 + 120}{2 ( - 3 )}, x_{2} = \frac{- 120 - 120}{2 ( - 3 )}

x = \frac{- 120 + 120}{2 ( - 3 )} : 0

x = \frac{- 120 - 120}{2 ( - 3 )} : 40

The solutions to the quadratic equation are:

x = 0, x = 40

12 x^{2} - 360 x + 2000 = 0

x, 3 x^{2} - x^{2} y + y^{4} = 1

x, 9 x^{2} + 4 y^{2} + z^{2} = 36

z^{2} - 12 z + 6 = 0

0 = 5 x^{2} - 3 x + 1

What are the solutions to the equation -0.3x^2+12x=0 ?
The solutions to the equation -0.3x^2+12x=0 are x=0,x=40
Find the zeros of -0.3x^2+12x=0
The zeros of -0.3x^2+12x=0 are x=0,x=40