What are the solutions to the equation 0=48t-16t^2 ?

The solutions to the equation 0=48t-16t^2 are t=0,t=3

Find the zeros of 0=48t-16t^2

The zeros of 0=48t-16t^2 are t=0,t=3

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0 = 48 t - 16 t^{2}

t = 0, t = 3

Solution steps

0 = 48 t - 16 t^{2}

Switch sides

48 t - 16 t^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

t_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

4 8^{2} - 4 (- 16) \cdot 0 = 48

t_{1, 2} = \frac{- 48 \pm 48}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 48 + 48}{2 ( - 16 )}, t_{2} = \frac{- 48 - 48}{2 ( - 16 )}

t = \frac{- 48 + 48}{2 ( - 16 )} : 0

t = \frac{- 48 - 48}{2 ( - 16 )} : 3

The solutions to the quadratic equation are:

t = 0, t = 3

30 x^{2} - 28 x + 6 = 0

\frac{4 a}{3 b c} - \frac{15 b}{5 a c}

x^{2} + 2 x + 26 = 0

- 5 v + 10 = - 7 v + 16

\frac{8 ^{- 5}}{8 ^{- 4}}

What are the solutions to the equation 0=48t-16t^2 ?
The solutions to the equation 0=48t-16t^2 are t=0,t=3
Find the zeros of 0=48t-16t^2
The zeros of 0=48t-16t^2 are t=0,t=3