What are the solutions to the equation 12x^2+4x+12=0 ?

The solutions to the equation 12x^2+4x+12=0 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6

Find the zeros of 12x^2+4x+12=0

The zeros of 12x^2+4x+12=0 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6

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12 x^{2} + 4 x + 12 = 0

x = - \frac{1}{6} + i \frac{35}{6}, x = - \frac{1}{6} - i \frac{35}{6}

Solution steps

12 x^{2} + 4 x + 12 = 0

Solve with the quadratic formula

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

Simplify

4^{2} - 4 \cdot 12 \cdot 12 : 435 i

x_{1, 2} = \frac{- 4 \pm 4 35 i}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 4 + 4 35 i}{2 \cdot 12}, x_{2} = \frac{- 4 - 4 35 i}{2 \cdot 12}

x = \frac{- 4 + 4 35 i}{2 \cdot 12} : - \frac{1}{6} + i \frac{35}{6}

x = \frac{- 4 - 4 35 i}{2 \cdot 12} : - \frac{1}{6} - i \frac{35}{6}

The solutions to the quadratic equation are:

x = - \frac{1}{6} + i \frac{35}{6}, x = - \frac{1}{6} - i \frac{35}{6}

5 t^{2} - 6 t + 600 = 0

x - 1 = 7 (x - 5) - (6 x - 4)

∣ 10 - z ∣ = 5

y^{- 4} \cdot y^{3}

- 3 (2 x + 1) > - 2 (x + 4)

What are the solutions to the equation 12x^2+4x+12=0 ?
The solutions to the equation 12x^2+4x+12=0 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6
Find the zeros of 12x^2+4x+12=0
The zeros of 12x^2+4x+12=0 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6