What is x^2-|3x+2|+x<= 0 ?

The solution to x^2-|3x+2|+x<= 0 is -sqrt(2)-2<= x<= sqrt(3)+1

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x^2-|3x+2|+x<= 0

x^{2} - ∣ 3 x + 2 ∣ + x \leq 0

- 2 - 2 \leq x \leq 3 + 1

Solution steps

x^{2} - ∣ 3 x + 2 ∣ + x \leq 0

Find positive and negative intervals

x < - \frac{2}{3}, x \geq - \frac{2}{3}

Solve the inequality for each interval

- 2 - 2 \leq x < - \frac{2}{3} or - \frac{2}{3} \leq x \leq 3 + 1

Merge Overlapping Intervals

- 2 - 2 \leq x \leq 3 + 1

∣ x - 11 ∣ + ∣ x + 2 ∣ \leq \frac{1}{4}

x^{2} - 4 > x^{2} - 4

x^{2} - 5 ∣ x ∣ + 6 \geq 0

x^{2} - 4 ∣ x ∣ + 3 < 0

∣ 4 (x - 1) + 10 ∣ \leq 22

What is x^2-|3x+2|+x<= 0 ?
The solution to x^2-|3x+2|+x<= 0 is -sqrt(2)-2<= x<= sqrt(3)+1