What are the solutions to the equation -6x=3x^2+1 ?

The solutions to the equation -6x=3x^2+1 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3

Find the zeros of -6x=3x^2+1

The zeros of -6x=3x^2+1 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3

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- 6 x = 3 x^{2} + 1

x = \frac{- 3 + 6}{3}, x = - \frac{3 + 6}{3}

Solution steps

- 6 x = 3 x^{2} + 1

Switch sides

3 x^{2} + 1 = - 6 x

Move

6 x

to the left side

3 x^{2} + 1 + 6 x = 0

Write in the standard form

a x^{2} + b x + c = 0

3 x^{2} + 6 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 \cdot 3 \cdot 1}{2 \cdot 3}

6^{2} - 4 \cdot 3 \cdot 1 = 26

x_{1, 2} = \frac{- 6 \pm 2 6}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 6 + 2 6}{2 \cdot 3}, x_{2} = \frac{- 6 - 2 6}{2 \cdot 3}

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

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

The solutions to the quadratic equation are:

x = \frac{- 3 + 6}{3}, x = - \frac{3 + 6}{3}

\frac{9}{- 7}

x^{2} = 324

225 x^{2} - 196 y^{2}

x^{3} - 4 x^{2} - 3 x + 12

- 7 = \frac{z}{2} + 1

What are the solutions to the equation -6x=3x^2+1 ?
The solutions to the equation -6x=3x^2+1 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3
Find the zeros of -6x=3x^2+1
The zeros of -6x=3x^2+1 are x=(-3+sqrt(6))/3 ,x=-(3+sqrt(6))/3