What are the solutions to the equation -0.6t^2+48t=0 ?

The solutions to the equation -0.6t^2+48t=0 are t=0,t=80

Find the zeros of -0.6t^2+48t=0

The zeros of -0.6t^2+48t=0 are t=0,t=80

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- 0.6 t^{2} + 48 t = 0

t = 0, t = 80

Solution steps

- 0.6 t^{2} + 48 t = 0

Multiply both sides by

10

- 6 t^{2} + 480 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 480 \pm 48 0 ^{2} - 4 ( - 6 ) \cdot 0}{2 ( - 6 )}

48 0^{2} - 4 (- 6) \cdot 0 = 480

t_{1, 2} = \frac{- 480 \pm 480}{2 ( - 6 )}

Separate the solutions

t_{1} = \frac{- 480 + 480}{2 ( - 6 )}, t_{2} = \frac{- 480 - 480}{2 ( - 6 )}

t = \frac{- 480 + 480}{2 ( - 6 )} : 0

t = \frac{- 480 - 480}{2 ( - 6 )} : 80

The solutions to the quadratic equation are:

t = 0, t = 80

2 z^{2} + 3 z - 6 = 0

3 y^{2} = 0

x^{2} = 1 0^{2} + 1 2^{2}

- 3 x^{2} + 195 x - 650 = 2200

x^{2} + 6 x - 144 = 0

What are the solutions to the equation -0.6t^2+48t=0 ?
The solutions to the equation -0.6t^2+48t=0 are t=0,t=80
Find the zeros of -0.6t^2+48t=0
The zeros of -0.6t^2+48t=0 are t=0,t=80