What is 10-3x^2>x^2+10-x ?

The solution to 10-3x^2>x^2+10-x is 0<x< 1/4

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10-3x^2>x^2+10-x

10 - 3 x^{2} > x^{2} + 10 - x

0 < x < \frac{1}{4}

Solution steps

10 - 3 x^{2} > x^{2} + 10 - x

Rewrite in standard form

- 4 x^{2} + x > 0

Factor

- 4 x^{2} + x : - x (4 x - 1)

- x (4 x - 1) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- x (4 x - 1)) (- 1) < 0 \cdot (- 1)

Simplify

x (4 x - 1) < 0

Identify the intervals

0 < x < \frac{1}{4}

C = \frac{5}{9} (50 - 32)

(4 x)^{2}

12 (- \frac{7}{3}) - 4 (- \frac{7}{3}) + 14

3 x + 2 < 14

2 b^{2} + 4 b - 70