What are the solutions to the equation 12x^2+x-6=0 ?

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

Find the zeros of 12x^2+x-6=0

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

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12 x^{2} + x - 6 = 0

x = \frac{2}{3}, x = - \frac{3}{4}

Solution steps

12 x^{2} + x - 6 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 \cdot 12 ( - 6 )}{2 \cdot 12}

1^{2} - 4 \cdot 12 (- 6) = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 \cdot 12}, x_{2} = \frac{- 1 - 17}{2 \cdot 12}

x = \frac{- 1 + 17}{2 \cdot 12} : \frac{2}{3}

x = \frac{- 1 - 17}{2 \cdot 12} : - \frac{3}{4}

The solutions to the quadratic equation are:

x = \frac{2}{3}, x = - \frac{3}{4}

9 (2 m - 3)^{2} + 8 = 449

(3 + 2 i) (3 - 2 i)

\frac{x + 1}{2} = \frac{x}{7}

6 x y - 3 x - 8 y + 4

p - 5 < 3

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