What are the solutions to the equation 320=144t-16t^2 ?

The solutions to the equation 320=144t-16t^2 are t=4,t=5

Find the zeros of 320=144t-16t^2

The zeros of 320=144t-16t^2 are t=4,t=5

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320 = 144 t - 16 t^{2}

t = 4, t = 5

Solution steps

320 = 144 t - 16 t^{2}

Switch sides

144 t - 16 t^{2} = 320

Move

320

to the left side

144 t - 16 t^{2} - 320 = 0

Write in the standard form

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

- 16 t^{2} + 144 t - 320 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 144 \pm 14 4 ^{2} - 4 ( - 16 ) ( - 320 )}{2 ( - 16 )}

14 4^{2} - 4 (- 16) (- 320) = 16

t_{1, 2} = \frac{- 144 \pm 16}{2 ( - 16 )}

Separate the solutions

t_{1} = \frac{- 144 + 16}{2 ( - 16 )}, t_{2} = \frac{- 144 - 16}{2 ( - 16 )}

t = \frac{- 144 + 16}{2 ( - 16 )} : 4

t = \frac{- 144 - 16}{2 ( - 16 )} : 5

The solutions to the quadratic equation are:

t = 4, t = 5

200 (\frac{( \frac{1 + 0.07}{12} ) ^{12 \cdot 35} - 1}{\frac{0.07}{12}})

x^{2} - 12 x - 1 = 0

0 = x^{2} + 2 x

- 44 > - 2 x - 8 \geq - 8

9 e + 4 = 5 e + 14 + 13 e

What are the solutions to the equation 320=144t-16t^2 ?
The solutions to the equation 320=144t-16t^2 are t=4,t=5
Find the zeros of 320=144t-16t^2
The zeros of 320=144t-16t^2 are t=4,t=5