What are the solutions to the equation 6x^2-3x-4=0 ?

The solutions to the equation 6x^2-3x-4=0 are x=(3+sqrt(105))/(12),x=(3-sqrt(105))/(12)

Find the zeros of 6x^2-3x-4=0

The zeros of 6x^2-3x-4=0 are x=(3+sqrt(105))/(12),x=(3-sqrt(105))/(12)

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

x = \frac{3 + 105}{12}, x = \frac{3 - 105}{12}

Decimal Notation

x = 1.10391 \dots, x = - 0.60391 \dots

Solution steps

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

Solve with the quadratic formula

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

(- 3)^{2} - 4 \cdot 6 (- 4) = 105

x_{1, 2} = \frac{- ( - 3 ) \pm 105}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 3 ) + 105}{2 \cdot 6}, x_{2} = \frac{- ( - 3 ) - 105}{2 \cdot 6}

x = \frac{- ( - 3 ) + 105}{2 \cdot 6} : \frac{3 + 105}{12}

x = \frac{- ( - 3 ) - 105}{2 \cdot 6} : \frac{3 - 105}{12}

The solutions to the quadratic equation are:

x = \frac{3 + 105}{12}, x = \frac{3 - 105}{12}

4 x^{2} - m^{2} n^{4} = 0

(z + 1) (z + 2) = 0

5 y = y^{2} - 6

x^{2} + 0.24 x - 0.0144 = 0

16 x^{2} + 8 x = 15

What are the solutions to the equation 6x^2-3x-4=0 ?
The solutions to the equation 6x^2-3x-4=0 are x=(3+sqrt(105))/(12),x=(3-sqrt(105))/(12)
Find the zeros of 6x^2-3x-4=0
The zeros of 6x^2-3x-4=0 are x=(3+sqrt(105))/(12),x=(3-sqrt(105))/(12)