What is -3x^2-2x+2>0 ?

The solution to -3x^2-2x+2>0 is (-sqrt(7)-1)/3 <x<(sqrt(7)-1)/3

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-3x^2-2x+2>0

- 3 x^{2} - 2 x + 2 > 0

\frac{- 7 - 1}{3} < x < \frac{7 - 1}{3}

Solution steps

- 3 x^{2} - 2 x + 2 > 0

Complete the square

- 3 x^{2} - 2 x + 2 : - 3 (x + \frac{1}{3})^{2} + \frac{7}{3}

- 3 (x + \frac{1}{3})^{2} + \frac{7}{3} > 0

Move

\frac{7}{3}

to the right side

- 3 (x + \frac{1}{3})^{2} > - \frac{7}{3}

Multiply both sides by

- 1

3 (x + \frac{1}{3})^{2} < \frac{7}{3}

Divide both sides by

3

(x + \frac{1}{3})^{2} < \frac{7}{9}

For

u^{n} < a

, if

n

is even then

- \frac{7}{9} < x + \frac{1}{3} < \frac{7}{9}

a < u < b

then

a < u and u < b

- \frac{7}{9} < x + \frac{1}{3} and x + \frac{1}{3} < \frac{7}{9}

- \frac{7}{9} < x + \frac{1}{3} : x > \frac{- 7 - 1}{3}

x + \frac{1}{3} < \frac{7}{9} : x < \frac{7 - 1}{3}

Combine the intervals

x > \frac{- 7 - 1}{3} and x < \frac{7 - 1}{3}

Merge Overlapping Intervals

\frac{- 7 - 1}{3} < x < \frac{7 - 1}{3}

\frac{5 x ^{4} y ^{5}}{5 x ^{2} y ^{8}}

b - 8 \leq - 2 b - 14

\frac{9 c ^{3}}{c ^{- 4}}

x^{2} + 11 x + 15 = - 3

- 2 p^{2} + 4000 p = 0

What is -3x^2-2x+2>0 ?
The solution to -3x^2-2x+2>0 is (-sqrt(7)-1)/3 <x<(sqrt(7)-1)/3