What is 2(2x^2-3x)>= 9 ?

The solution to 2(2x^2-3x)>= 9 is x = (3sqrt(5)+3)/4

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2(2x^2-3x)>= 9

2 (2 x^{2} - 3 x) \geq 9

x \leq \frac{- 3 5 + 3}{4} or x \geq \frac{3 5 + 3}{4}

Interval Notation

(- \infty, \frac{- 3 5 + 3}{4}] \cup [\frac{3 5 + 3}{4}, \infty)

Decimal Notation

x \leq - 0.92705 \dots or x \geq 2.42705 \dots

Solution steps

2 (2 x^{2} - 3 x) \geq 9

Rewrite in standard form

4 x^{2} - 6 x - 9 \geq 0

Complete the square

4 x^{2} - 6 x - 9 : 4 (x - \frac{3}{4})^{2} - \frac{45}{4}

4 (x - \frac{3}{4})^{2} - \frac{45}{4} \geq 0

Move

\frac{45}{4}

to the right side

4 (x - \frac{3}{4})^{2} \geq \frac{45}{4}

Divide both sides by

4

(x - \frac{3}{4})^{2} \geq \frac{45}{16}

For

u^{n} \geq a

, if

n

is even then

x - \frac{3}{4} \leq - \frac{45}{16} or x - \frac{3}{4} \geq \frac{45}{16}

x - \frac{3}{4} \leq - \frac{45}{16} : x \leq \frac{- 3 5 + 3}{4}

x - \frac{3}{4} \geq \frac{45}{16} : x \geq \frac{3 5 + 3}{4}

Combine the intervals

x \leq \frac{- 3 5 + 3}{4} or x \geq \frac{3 5 + 3}{4}

4 x - (7 x + 3) \leq 8 x + 2 (x - 3)

∣ 3 - 2 x ∣ < 5

- x^{2} - 2 x + 5 = 0

A = 9000 (1 + 8.75)^{15}

6 x^{2} + 11 x + 4 = 0

What is 2(2x^2-3x)>= 9 ?
The solution to 2(2x^2-3x)>= 9 is x<= (-3sqrt(5)+3)/4 \lor x>= (3sqrt(5)+3)/4