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

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

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

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

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48 t - 16 t^{2} = 0

t = 0, t = 3

Solution steps

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

\frac{- 5}{2 x - 4} > 0

(2 y - 3) (y - 2) = - 12 y + 18

- (\frac{5}{8})^{0}

2 - x^{2} = ∣ x ∣

3 x + 12 = 14 x + 6

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