What is -2x^2-3x+4<= 2 ?

The solution to -2x^2-3x+4 = 1/2

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-2x^2-3x+4<= 2

- 2 x^{2} - 3 x + 4 \leq 2

x \leq - 2 or x \geq \frac{1}{2}

Interval Notation

(- \infty, - 2] \cup [\frac{1}{2}, \infty)

Decimal Notation

x \leq - 2 or x \geq 0.5

Solution steps

- 2 x^{2} - 3 x + 4 \leq 2

Rewrite in standard form

- 2 x^{2} - 3 x + 2 \leq 0

Factor

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

- (2 x - 1) (x + 2) \leq 0

Multiply both sides by

- 1

(reverse the inequality)

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

Simplify

(2 x - 1) (x + 2) \geq 0

Identify the intervals

x \leq - 2 or x \geq \frac{1}{2}

\frac{6}{7} \cdot \frac{6}{7}

\frac{7}{10} (x + 8) = 18

a^{2} + 7^{2} = 1 2^{2}

\frac{( x ^{3} - 3 x y ^{2} + 2 y ^{3} )}{x + 2 y}

\frac{a + 1}{a ^{2} - 8 a + 15}