What are the solutions to the equation 4t^2+8t=0 ?

The solutions to the equation 4t^2+8t=0 are t=0,t=-2

Find the zeros of 4t^2+8t=0

The zeros of 4t^2+8t=0 are t=0,t=-2

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4 t^{2} + 8 t = 0

t = 0, t = - 2

Solution steps

4 t^{2} + 8 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 \cdot 4 \cdot 0}{2 \cdot 4}

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

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

Separate the solutions

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

t = \frac{- 8 + 8}{2 \cdot 4} : 0

t = \frac{- 8 - 8}{2 \cdot 4} : - 2

The solutions to the quadratic equation are:

t = 0, t = - 2

2 (3 x^{2} - 1) = 6 x + 5 x

144 x^{2} - 24 x + 1 = 0

0.1 x^{2} + 2.155 x + 20 = 200

12 x^{2} - 29 x - 8 = 0

2 x^{2} - 8 - 64 = 0

What are the solutions to the equation 4t^2+8t=0 ?
The solutions to the equation 4t^2+8t=0 are t=0,t=-2
Find the zeros of 4t^2+8t=0
The zeros of 4t^2+8t=0 are t=0,t=-2