What are the solutions to the equation 24t-0.48t^2=0 ?

The solutions to the equation 24t-0.48t^2=0 are t=0,t=50

Find the zeros of 24t-0.48t^2=0

The zeros of 24t-0.48t^2=0 are t=0,t=50

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24 t - 0.48 t^{2} = 0

t = 0, t = 50

Solution steps

24 t - 0.48 t^{2} = 0

Multiply both sides by

100

2400 t - 48 t^{2} = 0

Write in the standard form

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

- 48 t^{2} + 2400 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 2400 \pm 240 0 ^{2} - 4 ( - 48 ) \cdot 0}{2 ( - 48 )}

240 0^{2} - 4 (- 48) \cdot 0 = 2400

t_{1, 2} = \frac{- 2400 \pm 2400}{2 ( - 48 )}

Separate the solutions

t_{1} = \frac{- 2400 + 2400}{2 ( - 48 )}, t_{2} = \frac{- 2400 - 2400}{2 ( - 48 )}

t = \frac{- 2400 + 2400}{2 ( - 48 )} : 0

t = \frac{- 2400 - 2400}{2 ( - 48 )} : 50

The solutions to the quadratic equation are:

t = 0, t = 50

\frac{3}{4} x [\frac{2}{7} + \frac{5}{8}] = \frac{3}{4} \frac{x ^{2}}{7}

17 x^{2} - 40 x + 8 = 0

9 x^{2} + 8 (k - 5) x + 64 = 0

t, s = v_{0} \cdot t + a \cdot t^{2}

y, 9 y^{2} = x \cdot (3 - x)^{2}

What are the solutions to the equation 24t-0.48t^2=0 ?
The solutions to the equation 24t-0.48t^2=0 are t=0,t=50
Find the zeros of 24t-0.48t^2=0
The zeros of 24t-0.48t^2=0 are t=0,t=50