What is the answer to |x^2+3x-1|=2-x ?

The answer to |x^2+3x-1|=2-x is x=-2-sqrt(7)\lor x=-1\lor x=-2+sqrt(7)

Solve for x: |x^2+3x-1|=2-x

The solution is x=-2-sqrt(7)\lor x=-1\lor x=-2+sqrt(7)

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x^{2} + 3 x - 1 = 2 - x

x = - 2 - 7 or x = - 1 or x = - 2 + 7

Decimal

x = - 4.64575 \dots, x = - 1, x = 0.64575 \dots

Solution steps

x^{2} + 3 x - 1 = 2 - x

Find positive and negative intervals

x \leq \frac{- 13 - 3}{2}, \frac{- 13 - 3}{2} < x < \frac{13 - 3}{2}, x \geq \frac{13 - 3}{2}

Solve the inequality for each interval

x = - 2 - 7 or x = - 1 or x = - 2 + 7

∣ X + 5 ∣ = - 15

∣ 2 x - 11 ∣ = 6

x^{2} = x^{2} - 5 x + 3

2 ∣ z ∣ = ∣ z - 2 ∣

5 - ∣ r + 1 ∣ = 9

What is the answer to |x^2+3x-1|=2-x ?
The answer to |x^2+3x-1|=2-x is x=-2-sqrt(7)\lor x=-1\lor x=-2+sqrt(7)
Solve for x: |x^2+3x-1|=2-x
The solution is x=-2-sqrt(7)\lor x=-1\lor x=-2+sqrt(7)