What are the solutions to the equation 250=40t-0.8t^2 ?

The solutions to the equation 250=40t-0.8t^2 are t=(25(2-sqrt(2)))/2 ,t=(25(2+sqrt(2)))/2

Find the zeros of 250=40t-0.8t^2

The zeros of 250=40t-0.8t^2 are t=(25(2-sqrt(2)))/2 ,t=(25(2+sqrt(2)))/2

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250 = 40 t - 0.8 t^{2}

t = \frac{25 ( 2 - 2 )}{2}, t = \frac{25 ( 2 + 2 )}{2}

Decimal Notation

t = 7.32233 \dots, t = 42.67766 \dots

Solution steps

250 = 40 t - 0.8 t^{2}

Multiply both sides by

10

2500 = 400 t - 8 t^{2}

Switch sides

400 t - 8 t^{2} = 2500

Move

2500

to the left side

400 t - 8 t^{2} - 2500 = 0

Write in the standard form

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

- 8 t^{2} + 400 t - 2500 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 400 \pm 40 0 ^{2} - 4 ( - 8 ) ( - 2500 )}{2 ( - 8 )}

40 0^{2} - 4 (- 8) (- 2500) = 2002

t_{1, 2} = \frac{- 400 \pm 200 2}{2 ( - 8 )}

Separate the solutions

t_{1} = \frac{- 400 + 200 2}{2 ( - 8 )}, t_{2} = \frac{- 400 - 200 2}{2 ( - 8 )}

t = \frac{- 400 + 200 2}{2 ( - 8 )} : \frac{25 ( 2 - 2 )}{2}

t = \frac{- 400 - 200 2}{2 ( - 8 )} : \frac{25 ( 2 + 2 )}{2}

The solutions to the quadratic equation are:

t = \frac{25 ( 2 - 2 )}{2}, t = \frac{25 ( 2 + 2 )}{2}

3 x^{2} = 6 x

44

\frac{8}{6}

d (z - 2) + 5 (z - 2)

- 3 (- 1)^{2}

What are the solutions to the equation 250=40t-0.8t^2 ?
The solutions to the equation 250=40t-0.8t^2 are t=(25(2-sqrt(2)))/2 ,t=(25(2+sqrt(2)))/2
Find the zeros of 250=40t-0.8t^2
The zeros of 250=40t-0.8t^2 are t=(25(2-sqrt(2)))/2 ,t=(25(2+sqrt(2)))/2