What are the solutions to the equation 3x^2-x-8=0 ?

The solutions to the equation 3x^2-x-8=0 are x=(1+sqrt(97))/6 ,x=(1-sqrt(97))/6

Find the zeros of 3x^2-x-8=0

The zeros of 3x^2-x-8=0 are x=(1+sqrt(97))/6 ,x=(1-sqrt(97))/6

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3 x^{2} - x - 8 = 0

x = \frac{1 + 97}{6}, x = \frac{1 - 97}{6}

Decimal Notation

x = 1.80814 \dots, x = - 1.47480 \dots

Solution steps

3 x^{2} - x - 8 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 \cdot 3 ( - 8 )}{2 \cdot 3}

(- 1)^{2} - 4 \cdot 3 (- 8) = 97

x_{1, 2} = \frac{- ( - 1 ) \pm 97}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 97}{2 \cdot 3}, x_{2} = \frac{- ( - 1 ) - 97}{2 \cdot 3}

x = \frac{- ( - 1 ) + 97}{2 \cdot 3} : \frac{1 + 97}{6}

x = \frac{- ( - 1 ) - 97}{2 \cdot 3} : \frac{1 - 97}{6}

The solutions to the quadratic equation are:

x = \frac{1 + 97}{6}, x = \frac{1 - 97}{6}

- 8 x = 20

n^{2} - 3 n - 28 = 0

16 x^{4} - 49 x^{2} y^{2} + 25 x y

- \frac{5}{6} - \frac{1}{6}

(- \frac{11}{6}) + m = - \frac{2}{9}

What are the solutions to the equation 3x^2-x-8=0 ?
The solutions to the equation 3x^2-x-8=0 are x=(1+sqrt(97))/6 ,x=(1-sqrt(97))/6
Find the zeros of 3x^2-x-8=0
The zeros of 3x^2-x-8=0 are x=(1+sqrt(97))/6 ,x=(1-sqrt(97))/6