What are the solutions to the equation 3x^2-2x-10=0 ?

The solutions to the equation 3x^2-2x-10=0 are x=(1+sqrt(31))/3 ,x=(1-sqrt(31))/3

Find the zeros of 3x^2-2x-10=0

The zeros of 3x^2-2x-10=0 are x=(1+sqrt(31))/3 ,x=(1-sqrt(31))/3

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3 x^{2} - 2 x - 10 = 0

x = \frac{1 + 31}{3}, x = \frac{1 - 31}{3}

Decimal Notation

x = 2.18925 \dots, x = - 1.52258 \dots

Solution steps

3 x^{2} - 2 x - 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 \cdot 3 ( - 10 )}{2 \cdot 3}

(- 2)^{2} - 4 \cdot 3 (- 10) = 231

x_{1, 2} = \frac{- ( - 2 ) \pm 2 31}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 2 31}{2 \cdot 3}, x_{2} = \frac{- ( - 2 ) - 2 31}{2 \cdot 3}

x = \frac{- ( - 2 ) + 2 31}{2 \cdot 3} : \frac{1 + 31}{3}

x = \frac{- ( - 2 ) - 2 31}{2 \cdot 3} : \frac{1 - 31}{3}

The solutions to the quadratic equation are:

x = \frac{1 + 31}{3}, x = \frac{1 - 31}{3}

m^{2} - 5 mn - 36 n^{2}

51 2^{\frac{2}{3}}

(2 x - 5) (x + 3) \geq 0

- \frac{1}{24} + \frac{1}{160} + \frac{1}{3} - \frac{1}{5}

- 4 x + 3 = 20 x

What are the solutions to the equation 3x^2-2x-10=0 ?
The solutions to the equation 3x^2-2x-10=0 are x=(1+sqrt(31))/3 ,x=(1-sqrt(31))/3
Find the zeros of 3x^2-2x-10=0
The zeros of 3x^2-2x-10=0 are x=(1+sqrt(31))/3 ,x=(1-sqrt(31))/3