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

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

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

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

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

j = - 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

j = - 1

The solution to the quadratic equation is:

j = - 1

- 3 x^{2} - 4 x + 175 = 0

x, x^{2} - 4 x = y^{2} - 4 y

x^{2} + 5 x - 36 = 0

2 x - 5 = - x^{2}

4 (x^{2} - 6 x) = 24

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