What are the solutions to the equation 6x^2-3x-1=0 ?

The solutions to the equation 6x^2-3x-1=0 are x=(3+sqrt(33))/(12),x=(3-sqrt(33))/(12)

Find the zeros of 6x^2-3x-1=0

The zeros of 6x^2-3x-1=0 are x=(3+sqrt(33))/(12),x=(3-sqrt(33))/(12)

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6 x^{2} - 3 x - 1 = 0

x = \frac{3 + 33}{12}, x = \frac{3 - 33}{12}

Decimal Notation

x = 0.72871 \dots, x = - 0.22871 \dots

Solution steps

6 x^{2} - 3 x - 1 = 0

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 6 (- 1) = 33

x_{1, 2} = \frac{- ( - 3 ) \pm 33}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 33}{2 \cdot 6}, x_{2} = \frac{- ( - 3 ) - 33}{2 \cdot 6}

x = \frac{- ( - 3 ) + 33}{2 \cdot 6} : \frac{3 + 33}{12}

x = \frac{- ( - 3 ) - 33}{2 \cdot 6} : \frac{3 - 33}{12}

The solutions to the quadratic equation are:

x = \frac{3 + 33}{12}, x = \frac{3 - 33}{12}

- 6 x^{2} + 6 x

5 x^{4} + 10 x^{3} - 175 x^{2}

4 a + 4 b

\frac{x ^{2} + 4}{25 x ^{2} - 36} \geq 0

∣ 4 - 2 x ∣ + 8 = 3

What are the solutions to the equation 6x^2-3x-1=0 ?
The solutions to the equation 6x^2-3x-1=0 are x=(3+sqrt(33))/(12),x=(3-sqrt(33))/(12)
Find the zeros of 6x^2-3x-1=0
The zeros of 6x^2-3x-1=0 are x=(3+sqrt(33))/(12),x=(3-sqrt(33))/(12)