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

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

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

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

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

t = - 1

Solution steps

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

Solve with the quadratic formula

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

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

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

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

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

t = - 1

The solution to the quadratic equation is:

t = - 1

x^{2} - 2 x - 22 = 2 x - 1

lo g_{10} (9) = x^{2} + 3

x, x^{2} + 3 y^{2} + 2 x y = 0

a, a^{2} + 3 a d - d^{2} = 0

2 + 5 r - 10 r^{2} = 0

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