What are the solutions to the equation 0=448t-32t^2 ?

The solutions to the equation 0=448t-32t^2 are t=0,t=14

Find the zeros of 0=448t-32t^2

The zeros of 0=448t-32t^2 are t=0,t=14

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0 = 448 t - 32 t^{2}

t = 0, t = 14

Solution steps

0 = 448 t - 32 t^{2}

Switch sides

448 t - 32 t^{2} = 0

Write in the standard form

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

- 32 t^{2} + 448 t = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 448 \pm 44 8 ^{2} - 4 ( - 32 ) \cdot 0}{2 ( - 32 )}

44 8^{2} - 4 (- 32) \cdot 0 = 448

t_{1, 2} = \frac{- 448 \pm 448}{2 ( - 32 )}

Separate the solutions

t_{1} = \frac{- 448 + 448}{2 ( - 32 )}, t_{2} = \frac{- 448 - 448}{2 ( - 32 )}

t = \frac{- 448 + 448}{2 ( - 32 )} : 0

t = \frac{- 448 - 448}{2 ( - 32 )} : 14

The solutions to the quadratic equation are:

t = 0, t = 14

40 = 16 - 8 r

2 x + 5 + 2 x + 3 x = 7 x + 7

\frac{4}{64}

\frac{3 - 6}{1.25}

x^{2} + 5 x - 2 < 0

What are the solutions to the equation 0=448t-32t^2 ?
The solutions to the equation 0=448t-32t^2 are t=0,t=14
Find the zeros of 0=448t-32t^2
The zeros of 0=448t-32t^2 are t=0,t=14