What are the solutions to the equation 32t-0.8t^2=160 ?

The solutions to the equation 32t-0.8t^2=160 are t=10(2-sqrt(2)),t=10(2+sqrt(2))

Find the zeros of 32t-0.8t^2=160

The zeros of 32t-0.8t^2=160 are t=10(2-sqrt(2)),t=10(2+sqrt(2))

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32 t - 0.8 t^{2} = 160

t = 10 (2 - 2), t = 10 (2 + 2)

Solution steps

32 t - 0.8 t^{2} = 160

Multiply both sides by

10

320 t - 8 t^{2} = 1600

Move

1600

to the left side

320 t - 8 t^{2} - 1600 = 0

Write in the standard form

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

- 8 t^{2} + 320 t - 1600 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 320 \pm 32 0 ^{2} - 4 ( - 8 ) ( - 1600 )}{2 ( - 8 )}

32 0^{2} - 4 (- 8) (- 1600) = 1602

t_{1, 2} = \frac{- 320 \pm 160 2}{2 ( - 8 )}

Separate the solutions

t_{1} = \frac{- 320 + 160 2}{2 ( - 8 )}, t_{2} = \frac{- 320 - 160 2}{2 ( - 8 )}

t = \frac{- 320 + 160 2}{2 ( - 8 )} : 10 (2 - 2)

t = \frac{- 320 - 160 2}{2 ( - 8 )} : 10 (2 + 2)

The solutions to the quadratic equation are:

t = 10 (2 - 2), t = 10 (2 + 2)

2 x^{2} - 5 x + 4 = 0

8 - 11

\frac{\frac{5}{x} + \frac{1}{2 y}}{\frac{1}{2}}

x^{2} - 5 x - 24

x - 2 = \frac{3}{4} (x + 4)

What are the solutions to the equation 32t-0.8t^2=160 ?
The solutions to the equation 32t-0.8t^2=160 are t=10(2-sqrt(2)),t=10(2+sqrt(2))
Find the zeros of 32t-0.8t^2=160
The zeros of 32t-0.8t^2=160 are t=10(2-sqrt(2)),t=10(2+sqrt(2))