What are the solutions to the equation -6x^2+2x-4=9x ?

The solutions to the equation -6x^2+2x-4=9x are x=-7/12-i(sqrt(47))/(12),x=-7/12+i(sqrt(47))/(12)

Find the zeros of -6x^2+2x-4=9x

The zeros of -6x^2+2x-4=9x are x=-7/12-i(sqrt(47))/(12),x=-7/12+i(sqrt(47))/(12)

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- 6 x^{2} + 2 x - 4 = 9 x

x = - \frac{7}{12} - i \frac{47}{12}, x = - \frac{7}{12} + i \frac{47}{12}

Solution steps

- 6 x^{2} + 2 x - 4 = 9 x

Move

9 x

to the left side

- 6 x^{2} - 7 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 6 ) ( - 4 )}{2 ( - 6 )}

Simplify

(- 7)^{2} - 4 (- 6) (- 4) : 47 i

x_{1, 2} = \frac{- ( - 7 ) \pm 47 i}{2 ( - 6 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 47 i}{2 ( - 6 )}, x_{2} = \frac{- ( - 7 ) - 47 i}{2 ( - 6 )}

x = \frac{- ( - 7 ) + 47 i}{2 ( - 6 )} : - \frac{7}{12} - i \frac{47}{12}

x = \frac{- ( - 7 ) - 47 i}{2 ( - 6 )} : - \frac{7}{12} + i \frac{47}{12}

The solutions to the quadratic equation are:

x = - \frac{7}{12} - i \frac{47}{12}, x = - \frac{7}{12} + i \frac{47}{12}

(x + 3)^{2} = (x + 1)^{2} + (2 x)^{2}

3 x^{2} + 3 x - 9 = 0

4 r^{2} + r + 2 = 0

x^{2} + 14 - x = - 10 x

x, x^{2} + 8 x - 9 = 0

What are the solutions to the equation -6x^2+2x-4=9x ?
The solutions to the equation -6x^2+2x-4=9x are x=-7/12-i(sqrt(47))/(12),x=-7/12+i(sqrt(47))/(12)
Find the zeros of -6x^2+2x-4=9x
The zeros of -6x^2+2x-4=9x are x=-7/12-i(sqrt(47))/(12),x=-7/12+i(sqrt(47))/(12)