What are the solutions to the equation 3x^2-9x+1=0 ?

The solutions to the equation 3x^2-9x+1=0 are x=(9+sqrt(69))/6 ,x=(9-sqrt(69))/6

Find the zeros of 3x^2-9x+1=0

The zeros of 3x^2-9x+1=0 are x=(9+sqrt(69))/6 ,x=(9-sqrt(69))/6

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3 x^{2} - 9 x + 1 = 0

x = \frac{9 + 69}{6}, x = \frac{9 - 69}{6}

Decimal Notation

x = 2.88443 \dots, x = 0.11556 \dots

Solution steps

3 x^{2} - 9 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 9 ) \pm ( - 9 ) ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

(- 9)^{2} - 4 \cdot 3 \cdot 1 = 69

x_{1, 2} = \frac{- ( - 9 ) \pm 69}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 9 ) + 69}{2 \cdot 3}, x_{2} = \frac{- ( - 9 ) - 69}{2 \cdot 3}

x = \frac{- ( - 9 ) + 69}{2 \cdot 3} : \frac{9 + 69}{6}

x = \frac{- ( - 9 ) - 69}{2 \cdot 3} : \frac{9 - 69}{6}

The solutions to the quadratic equation are:

x = \frac{9 + 69}{6}, x = \frac{9 - 69}{6}

5 ∣ 2 x - 1 ∣ + 1 = 6

3 v^{2} - 81 = 0

x^{2} y^{2}

0 - 8

3 s^{2} + 10 s + 6 = 30

What are the solutions to the equation 3x^2-9x+1=0 ?
The solutions to the equation 3x^2-9x+1=0 are x=(9+sqrt(69))/6 ,x=(9-sqrt(69))/6
Find the zeros of 3x^2-9x+1=0
The zeros of 3x^2-9x+1=0 are x=(9+sqrt(69))/6 ,x=(9-sqrt(69))/6