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

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

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

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

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12 x^{2} + x = 6

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

Decimal Notation

x = 0.66666 \dots, x = - 0.75

Solution steps

12 x^{2} + x = 6

Move

6

to the left side

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}

(x - 4)^{2} = 0

- (\frac{1}{x})^{2}

∣ 3 + 2 x ∣ < ∣ 4 - x ∣

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

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