What are the solutions to the equation solvefor y,y^2+2y+1=0 ?

The solutions to the equation solvefor y,y^2+2y+1=0 are y=-1

Find the zeros of solvefor y,y^2+2y+1=0

The zeros of solvefor y,y^2+2y+1=0 are y=-1

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y, y^{2} + 2 y + 1 = 0

y = - 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

y = - 1

The solution to the quadratic equation is:

y = - 1

t, A t - \frac{( B t ^{2} )}{2} = 0

(2 x - 1) (x + 2) = 8 x - 4

3 x^{2} - 26 = 1

x^{2} - 3 x - 19 = 0

(x - 4)^{2} - 10 = (x - 1)^{2} - x^{2}

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