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

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

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

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

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

D = - 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

D = - 1

The solution to the quadratic equation is:

D = - 1

2 d = a t^{2}

k^{2} - 4 k + 69 = - 3

x^{2} - 25 = 12, x + 5 = 4, x - 5

x^{2} - 10 x + 25 = - 16 + 25

\frac{- 1}{6} x^{2} + x - 3 = 0

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