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

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

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

x^{2} - ∣ 3 x + 2 ∣ + x \geq 0

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

Solution steps

x^{2} - ∣ 3 x + 2 ∣ + x \geq 0

Find positive and negative intervals

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

Solve the inequality for each interval

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

\frac{x + 5}{∣ 1 - x ∣} > 1

∣ 2 x - 10 ∣ > 10

∣ x ∣^{x^{2} - x - 2} < 1

∣ 5 x - 2 ∣ \leq ∣ 3 x + 5 ∣

∣ x + 3 ∣ > ∣ 2 x - 4 ∣

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