What is the answer to |x^2+4x-1|=7 ?

The answer to |x^2+4x-1|=7 is x=-2(1+sqrt(3))\lor x=2(sqrt(3)-1)

Solve for x: |x^2+4x-1|=7

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

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x^{2} + 4 x - 1 = 7

x = - 2 (1 + 3) or x = 2 (3 - 1)

Decimal Notation

x = - 5.46410 \dots, x = 1.46410 \dots

Solution steps

x^{2} + 4 x - 1 = 7

Apply absolute rule: If

∣ u ∣ = a, a > 0

then

u = a

u = - a

x^{2} + 4 x - 1 = - 7 or x^{2} + 4 x - 1 = 7

x^{2} + 4 x - 1 = - 7 or x^{2} + 4 x - 1 = 7

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

x = - 2 (1 + 3) or x = 2 (3 - 1)

∣ 6 x + 3 ∣ = 15

∣ 3 k - 6 ∣ = 2

∣ 8 + a ∣ = 10

∣ 2 x + 5 ∣ = 7 x - 2

∣ x - 4 ∣ = 17

What is the answer to |x^2+4x-1|=7 ?
The answer to |x^2+4x-1|=7 is x=-2(1+sqrt(3))\lor x=2(sqrt(3)-1)
Solve for x: |x^2+4x-1|=7
The solution is x=-2(1+sqrt(3))\lor x=2(sqrt(3)-1)