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

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

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

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

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6 x^{2} + x = 12

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

Solution steps

6 x^{2} + x = 12

Move

12

to the left side

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

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 \cdot 6}, x_{2} = \frac{- 1 - 17}{2 \cdot 6}

x = \frac{- 1 + 17}{2 \cdot 6} : \frac{4}{3}

x = \frac{- 1 - 17}{2 \cdot 6} : - \frac{3}{2}

The solutions to the quadratic equation are:

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

∣ x ∣ + 5 < 16

x^{2} - 10 x = 0

(4 x)^{3}

2 x^{2} + x^{2} = 5^{2}

2 ln (1)

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