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

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

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

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

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

v = - 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

v = - 1

The solution to the quadratic equation is:

v = - 1

10 u^{2} + 17 u + 3 = 0

\frac{7}{4} - \frac{2}{3}

14 + \frac{x}{2} = 3

x + \frac{2}{3} + \frac{1}{3} = 1

\frac{x + 3}{x + 2} < 0

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