What are the solutions to the equation 18x^2+4x+12=6x^2 ?

The solutions to the equation 18x^2+4x+12=6x^2 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6

Find the zeros of 18x^2+4x+12=6x^2

The zeros of 18x^2+4x+12=6x^2 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6

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18 x^{2} + 4 x + 12 = 6 x^{2}

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

Solution steps

18 x^{2} + 4 x + 12 = 6 x^{2}

Move

6 x^{2}

to the left side

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}

2 x + 0 = - 4

(x + y)^{3} - 64

t - 19 = 2

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

12 - (- 2)

What are the solutions to the equation 18x^2+4x+12=6x^2 ?
The solutions to the equation 18x^2+4x+12=6x^2 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6
Find the zeros of 18x^2+4x+12=6x^2
The zeros of 18x^2+4x+12=6x^2 are x=-1/6+i(sqrt(35))/6 ,x=-1/6-i(sqrt(35))/6