What are the solutions to the equation 6x^2-3x-5=0 ?

The solutions to the equation 6x^2-3x-5=0 are x=(3+sqrt(129))/(12),x=(3-sqrt(129))/(12)

Find the zeros of 6x^2-3x-5=0

The zeros of 6x^2-3x-5=0 are x=(3+sqrt(129))/(12),x=(3-sqrt(129))/(12)

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6 x^{2} - 3 x - 5 = 0

x = \frac{3 + 129}{12}, x = \frac{3 - 129}{12}

Solution steps

6 x^{2} - 3 x - 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 3 ) \pm ( - 3 ) ^{2} - 4 \cdot 6 ( - 5 )}{2 \cdot 6}

(- 3)^{2} - 4 \cdot 6 (- 5) = 129

x_{1, 2} = \frac{- ( - 3 ) \pm 129}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 129}{2 \cdot 6}, x_{2} = \frac{- ( - 3 ) - 129}{2 \cdot 6}

x = \frac{- ( - 3 ) + 129}{2 \cdot 6} : \frac{3 + 129}{12}

x = \frac{- ( - 3 ) - 129}{2 \cdot 6} : \frac{3 - 129}{12}

The solutions to the quadratic equation are:

x = \frac{3 + 129}{12}, x = \frac{3 - 129}{12}

27 x^{2} + 12 x - 8 = 0

- 3 x = - 15

14 x - 7 > 12 x

x^{4} - 11 x^{2} - 80

4 n^{2} - 1

What are the solutions to the equation 6x^2-3x-5=0 ?
The solutions to the equation 6x^2-3x-5=0 are x=(3+sqrt(129))/(12),x=(3-sqrt(129))/(12)
Find the zeros of 6x^2-3x-5=0
The zeros of 6x^2-3x-5=0 are x=(3+sqrt(129))/(12),x=(3-sqrt(129))/(12)