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

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

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

The solution is x=-2+sqrt(3)\lor x=1

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

x = - 2 + 3 or x = 1

Decimal

x = - 0.26794 \dots, x = 1

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

No Solution or x = - 2 + 3 or x = 1

x = - 2 + 3 or x = 1

5 ∣ x - 4 ∣ = 90

3 = ∣ x + 1 ∣ - ∣ x - 2 ∣

\frac{1}{3} ∣ 5 J + 2 ∣ + 11 = 2

∣ x - 1 ∣ = x^{2} - 1

∣ 6 n - 9 ∣ = 3

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