What are the solutions to the equation x=((2+12x^2))/(-11) ?

The solutions to the equation x=((2+12x^2))/(-11) are x=-2/3 ,x=-1/4

Find the zeros of x=((2+12x^2))/(-11)

The zeros of x=((2+12x^2))/(-11) are x=-2/3 ,x=-1/4

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x = \frac{( 2 + 12 x ^{2} )}{- 11}

x = - \frac{2}{3}, x = - \frac{1}{4}

Solution steps

x = \frac{( 2 + 12 x ^{2} )}{- 11}

Multiply both sides by

11

11 x = - (2 + 12 x^{2})

Expand

- (2 + 12 x^{2}) : - 2 - 12 x^{2}

11 x = - 2 - 12 x^{2}

Switch sides

- 2 - 12 x^{2} = 11 x

Move

11 x

to the left side

- 2 - 12 x^{2} - 11 x = 0

Write in the standard form

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

- 12 x^{2} - 11 x - 2 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 ( - 12 ) ( - 2 )}{2 ( - 12 )}

(- 11)^{2} - 4 (- 12) (- 2) = 5

x_{1, 2} = \frac{- ( - 11 ) \pm 5}{2 ( - 12 )}

Separate the solutions

x_{1} = \frac{- ( - 11 ) + 5}{2 ( - 12 )}, x_{2} = \frac{- ( - 11 ) - 5}{2 ( - 12 )}

x = \frac{- ( - 11 ) + 5}{2 ( - 12 )} : - \frac{2}{3}

x = \frac{- ( - 11 ) - 5}{2 ( - 12 )} : - \frac{1}{4}

The solutions to the quadratic equation are:

x = - \frac{2}{3}, x = - \frac{1}{4}

3 x + 4 (5 - 2 x) < \frac{x}{2}

\frac{3}{7} (35 x - 14) \leq \frac{21 x}{2} + 3

4 x^{2} + 36 = 0

- 6 x^{2} + 23 x + 4

x^{2} - 1 > x

What are the solutions to the equation x=((2+12x^2))/(-11) ?
The solutions to the equation x=((2+12x^2))/(-11) are x=-2/3 ,x=-1/4
Find the zeros of x=((2+12x^2))/(-11)
The zeros of x=((2+12x^2))/(-11) are x=-2/3 ,x=-1/4