What are the solutions to the equation 2x^2+4x+2=0 ?

The solutions to the equation 2x^2+4x+2=0 are x=-1

Find the zeros of 2x^2+4x+2=0

The zeros of 2x^2+4x+2=0 are x=-1

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2 x^{2} + 4 x + 2 = 0

x = - 1

Solution steps

2 x^{2} + 4 x + 2 = 0

Solve with the quadratic formula

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

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

x_{1, 2} = \frac{- 4 \pm 0}{2 \cdot 2}

x = \frac{- 4}{2 \cdot 2}

\frac{- 4}{2 \cdot 2} = - 1

x = - 1

The solution to the quadratic equation is:

x = - 1

- 3 < \frac{k}{3}

\frac{81}{16}

\frac{1}{2} - \frac{1}{k + 1}

∣ 7 m ∣ + 4 = 25

3 q^{2} - q - 14 = 0

What are the solutions to the equation 2x^2+4x+2=0 ?
The solutions to the equation 2x^2+4x+2=0 are x=-1
Find the zeros of 2x^2+4x+2=0
The zeros of 2x^2+4x+2=0 are x=-1