What is -2x^2+3x+12>10 ?

The solution to -2x^2+3x+12>10 is -1/2 <x<2

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-2x^2+3x+12>10

- 2 x^{2} + 3 x + 12 > 10

- \frac{1}{2} < x < 2

Interval Notation

(- \frac{1}{2}, 2)

Decimal Notation

- 0.5 < x < 2

Solution steps

- 2 x^{2} + 3 x + 12 > 10

Rewrite in standard form

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

Factor

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

- (2 x + 1) (x - 2) > 0

Multiply both sides by

- 1

(reverse the inequality)

(- (2 x + 1) (x - 2)) (- 1) < 0 \cdot (- 1)

Simplify

(2 x + 1) (x - 2) < 0

Identify the intervals

- \frac{1}{2} < x < 2

2 i - 7 i + 10

3 a + 10 = 4 a - 2

36 x^{2} - 64 = 0

(- 1)^{\frac{5}{3}}

\frac{x ^{3} - x ^{2} - 11 x + 3}{2 x ^{3} - 7 x ^{2} - 19 x + 60} \geq 0