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

The solutions to the equation -2x^2+12x=0 are x=0,x=6

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

The zeros of -2x^2+12x=0 are x=0,x=6

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

x = 0, x = 6

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

x = \frac{- 12 - 12}{2 ( - 2 )} : 6

The solutions to the quadratic equation are:

x = 0, x = 6

45 x^{2} - 9 x - 9

6 x^{2} - 7 x - 2 = 0

1 + 14 x^{2} y + 49 x^{4} y^{2}

48 = 10 w - 7 w

- \frac{1.95 E 26}{π ( 1.5 \cdot 1 0 ^{11} ) ^{3}}

What are the solutions to the equation -2x^2+12x=0 ?
The solutions to the equation -2x^2+12x=0 are x=0,x=6
Find the zeros of -2x^2+12x=0
The zeros of -2x^2+12x=0 are x=0,x=6