What are the solutions to the equation 3x^2-4x-13=0 ?

The solutions to the equation 3x^2-4x-13=0 are x=(2+sqrt(43))/3 ,x=(2-sqrt(43))/3

Find the zeros of 3x^2-4x-13=0

The zeros of 3x^2-4x-13=0 are x=(2+sqrt(43))/3 ,x=(2-sqrt(43))/3

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3 x^{2} - 4 x - 13 = 0

x = \frac{2 + 43}{3}, x = \frac{2 - 43}{3}

Decimal Notation

x = 2.85247 \dots, x = - 1.51914 \dots

Solution steps

3 x^{2} - 4 x - 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 ( - 13 )}{2 \cdot 3}

(- 4)^{2} - 4 \cdot 3 (- 13) = 243

x_{1, 2} = \frac{- ( - 4 ) \pm 2 43}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- ( - 4 ) + 2 43}{2 \cdot 3}, x_{2} = \frac{- ( - 4 ) - 2 43}{2 \cdot 3}

x = \frac{- ( - 4 ) + 2 43}{2 \cdot 3} : \frac{2 + 43}{3}

x = \frac{- ( - 4 ) - 2 43}{2 \cdot 3} : \frac{2 - 43}{3}

The solutions to the quadratic equation are:

x = \frac{2 + 43}{3}, x = \frac{2 - 43}{3}

6 x + 24 = - 36

- 3 - \frac{3}{5}

lo g_{2} (x + 1) + lo g_{x + 1} (2) \leq \frac{5}{2}

\frac{x + 1}{x + 6} > 0

\frac{12 - ( - 2 )}{7} + (- 3) + (2) (- 9)

What are the solutions to the equation 3x^2-4x-13=0 ?
The solutions to the equation 3x^2-4x-13=0 are x=(2+sqrt(43))/3 ,x=(2-sqrt(43))/3
Find the zeros of 3x^2-4x-13=0
The zeros of 3x^2-4x-13=0 are x=(2+sqrt(43))/3 ,x=(2-sqrt(43))/3