What are the solutions to the equation -3c^2+6c=0 ?

The solutions to the equation -3c^2+6c=0 are c=0,c=2

Find the zeros of -3c^2+6c=0

The zeros of -3c^2+6c=0 are c=0,c=2

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- 3 c^{2} + 6 c = 0

c = 0, c = 2

Solution steps

- 3 c^{2} + 6 c = 0

Solve with the quadratic formula

c_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 3 ) \cdot 0}{2 ( - 3 )}

6^{2} - 4 (- 3) \cdot 0 = 6

c_{1, 2} = \frac{- 6 \pm 6}{2 ( - 3 )}

Separate the solutions

c_{1} = \frac{- 6 + 6}{2 ( - 3 )}, c_{2} = \frac{- 6 - 6}{2 ( - 3 )}

c = \frac{- 6 + 6}{2 ( - 3 )} : 0

c = \frac{- 6 - 6}{2 ( - 3 )} : 2

The solutions to the quadratic equation are:

c = 0, c = 2

\frac{5}{2} x^{2} - 4 x + 2 = 0

x^{2} + 175 = 0

x^{2} + 15 x - 8 = 0

2 x (x - 5) + 2 = 3 x

5 x - 7 x^{2} = 7

What are the solutions to the equation -3c^2+6c=0 ?
The solutions to the equation -3c^2+6c=0 are c=0,c=2
Find the zeros of -3c^2+6c=0
The zeros of -3c^2+6c=0 are c=0,c=2