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

The solutions to the equation 1=x^2+2x-1 are x=-1+sqrt(3),x=-1-sqrt(3)

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

The zeros of 1=x^2+2x-1 are x=-1+sqrt(3),x=-1-sqrt(3)

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

x = - 1 + 3, x = - 1 - 3

Decimal Notation

x = 0.73205 \dots, x = - 2.73205 \dots

Solution steps

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

Switch sides

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

Move

1

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 2) = 23

x_{1, 2} = \frac{- 2 \pm 2 3}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 2 + 2 3}{2 \cdot 1}, x_{2} = \frac{- 2 - 2 3}{2 \cdot 1}

x = \frac{- 2 + 2 3}{2 \cdot 1} : - 1 + 3

x = \frac{- 2 - 2 3}{2 \cdot 1} : - 1 - 3

The solutions to the quadratic equation are:

x = - 1 + 3, x = - 1 - 3

x^{2} - 20 x + 75

\frac{5}{2} + 2

3 x^{2} + 19 x - 14 = 0

- \frac{1}{3} (3)

What are the solutions to the equation 1=x^2+2x-1 ?
The solutions to the equation 1=x^2+2x-1 are x=-1+sqrt(3),x=-1-sqrt(3)
Find the zeros of 1=x^2+2x-1
The zeros of 1=x^2+2x-1 are x=-1+sqrt(3),x=-1-sqrt(3)