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

The solutions to the equation 1x^2+2x+1=0 are x=-1

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

The zeros of 1x^2+2x+1=0 are x=-1

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

x = - 1

Solution steps

1 \cdot x^{2} + 2 x + 1 = 0

Refine

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 = 0

x_{1, 2} = \frac{- 2 \pm 0}{2 \cdot 1}

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

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

x = - 1

The solution to the quadratic equation is:

x = - 1

2 - x < 3 x + 1 < 10 x

8 y^{3} - 27

r s^{3} - 27 r t^{3} + 3 s^{3} - 81 t^{3}

\frac{g ^{8} h ^{2} m}{h g ^{7}}

- 6 (3 p - 3) = 90

What are the solutions to the equation 1x^2+2x+1=0 ?
The solutions to the equation 1x^2+2x+1=0 are x=-1
Find the zeros of 1x^2+2x+1=0
The zeros of 1x^2+2x+1=0 are x=-1