What is (1)2<= ((x^2-3x))/2 <5 ?

The solution to (1)2<= ((x^2-3x))/2 <5 is -2<x<=-1\lor 4<= x<5

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(1)2<= ((x^2-3x))/2 <5

(1) 2 \leq \frac{( x ^{2} - 3 x )}{2} < 5

- 2 < x \leq - 1 or 4 \leq x < 5

Interval Notation

(- 2, - 1] \cup [4, 5)

Solution steps

(1) \cdot 2 \leq \frac{( x ^{2} - 3 x )}{2} < 5

a \leq u < b

then

a \leq u and u < b

(1) \cdot 2 \leq \frac{( x ^{2} - 3 x )}{2} and \frac{( x ^{2} - 3 x )}{2} < 5

1 \cdot 2 \leq \frac{x ^{2} - 3 x}{2} : x \leq - 1 or x \geq 4

\frac{x ^{2} - 3 x}{2} < 5 : - 2 < x < 5

Combine the intervals

(x \leq - 1 or x \geq 4) and - 2 < x < 5

Merge Overlapping Intervals

- 2 < x \leq - 1 or 4 \leq x < 5

\frac{1}{2} < [x] < \frac{7}{8}

\frac{- 5}{3} < \frac{1}{2} - \frac{1}{3} x < \frac{- 3}{2}

3 < x + 3 < 6

- 5 < [∣ x ∣ - 7] < 5

- 12 < x < 13

What is (1)2<= ((x^2-3x))/2 <5 ?
The solution to (1)2<= ((x^2-3x))/2 <5 is -2<x<=-1\lor 4<= x<5