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

The solutions to the equation -3x^2+6x-2=0 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3

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

The zeros of -3x^2+6x-2=0 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3

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

x = \frac{3 - 3}{3}, x = \frac{3 + 3}{3}

Decimal Notation

x = 0.42264 \dots, x = 1.57735 \dots

Solution steps

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

Solve with the quadratic formula

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

6^{2} - 4 (- 3) (- 2) = 23

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = \frac{3 - 3}{3}, x = \frac{3 + 3}{3}

- 2 \leq - 2 x + 4 < 12

- 2 + (- 7)

\frac{1}{2} x^{2} + \frac{1}{6} x - 1 = 0

\frac{2}{5} + \frac{4}{3} \times \frac{3}{5} - (\frac{1}{4})^{2}

∣ x - 7 ∣ - 5 = 3

What are the solutions to the equation -3x^2+6x-2=0 ?
The solutions to the equation -3x^2+6x-2=0 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3
Find the zeros of -3x^2+6x-2=0
The zeros of -3x^2+6x-2=0 are x=(3-sqrt(3))/3 ,x=(3+sqrt(3))/3