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

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

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

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

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

s = - 1

Solution steps

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

Solve with the quadratic formula

s_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 1 \cdot 1}{2 \cdot 1}

2^{2} - 4 \cdot 1 \cdot 1 = 0

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

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

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

s = - 1

The solution to the quadratic equation is:

s = - 1

2 (4 x - 20 + 5 x) > 2 (3 x + 1) - 3 (2 x + 2)

\frac{y}{2} - 1 = \frac{y}{6} + 5

- 12 x^{2} + 72 x = 0

216 + y^{6}

- 3 d + 7 \geq - 5

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