What are the solutions to the equation 2t^2+8t+8=0 ?

The solutions to the equation 2t^2+8t+8=0 are t=-2

Find the zeros of 2t^2+8t+8=0

The zeros of 2t^2+8t+8=0 are t=-2

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2 t^{2} + 8 t + 8 = 0

t = - 2

Solution steps

2 t^{2} + 8 t + 8 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 2 \cdot 8}{2 \cdot 2}

8^{2} - 4 \cdot 2 \cdot 8 = 0

t_{1, 2} = \frac{- 8 \pm 0}{2 \cdot 2}

t = \frac{- 8}{2 \cdot 2}

\frac{- 8}{2 \cdot 2} = - 2

t = - 2

The solution to the quadratic equation is:

t = - 2

81 (x - 2)^{2} = 16

4 x^{2} - 6 x = 3

64 x^{2} + 40 x + 16 = 0

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

z, A = 4 w^{2} - 9 z^{2}

What are the solutions to the equation 2t^2+8t+8=0 ?
The solutions to the equation 2t^2+8t+8=0 are t=-2
Find the zeros of 2t^2+8t+8=0
The zeros of 2t^2+8t+8=0 are t=-2