What are the solutions to the equation 0=x^2+4x-1 ?

The solutions to the equation 0=x^2+4x-1 are x=-2+sqrt(5),x=-2-sqrt(5)

Find the zeros of 0=x^2+4x-1

The zeros of 0=x^2+4x-1 are x=-2+sqrt(5),x=-2-sqrt(5)

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

x = - 2 + 5, x = - 2 - 5

Solution steps

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

Switch sides

x^{2} + 4 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 1 \cdot ( - 1 )}{2 \cdot 1}

4^{2} - 4 \cdot 1 \cdot (- 1) = 25

x_{1, 2} = \frac{- 4 \pm 2 5}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 4 + 2 5}{2 \cdot 1}, x_{2} = \frac{- 4 - 2 5}{2 \cdot 1}

x = \frac{- 4 + 2 5}{2 \cdot 1} : - 2 + 5

x = \frac{- 4 - 2 5}{2 \cdot 1} : - 2 - 5

The solutions to the quadratic equation are:

x = - 2 + 5, x = - 2 - 5

∣ x - 1 ∣ \geq 3

2 x - 2 = x + 1

64 x - 24

- 5 + {2 (3 - 4) + 3 [(5 + 1) - 2]}

2 x^{2} + 6 x = - 18

What are the solutions to the equation 0=x^2+4x-1 ?
The solutions to the equation 0=x^2+4x-1 are x=-2+sqrt(5),x=-2-sqrt(5)
Find the zeros of 0=x^2+4x-1
The zeros of 0=x^2+4x-1 are x=-2+sqrt(5),x=-2-sqrt(5)