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

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

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

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

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160 = 32 t - 0.8 t^{2}

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

Decimal Notation

t = 5.85786 \dots, t = 34.14213 \dots

Solution steps

160 = 32 t - 0.8 t^{2}

Multiply both sides by

10

1600 = 320 t - 8 t^{2}

Switch sides

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)

4 y - 2 = 33 - 3 y

(8 x^{3})^{\frac{2}{3}}

x^{2} + 11 x + 6 = 0

- \frac{3 π}{2} + \frac{π}{2}

8 w + 52 = 164

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