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

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

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

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

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x^{2} + 2 x = 1

x = - 1 + 2, x = - 1 - 2

Solution steps

x^{2} + 2 x = 1

Move

1

to the left side

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

Solve with the quadratic formula

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

2^{2} - 4 \cdot 1 \cdot (- 1) = 22

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 + 2, x = - 1 - 2

\frac{j}{2} + 1 \leq - 1

2 x^{2} - 9 x = 1

2 = 3

x^{2} + 8 x + 37 = 0

2 s - 8 = 12

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