What are the solutions to the equation 200=t(18-0.2t) ?

The solutions to the equation 200=t(18-0.2t) are t=(18-2sqrt(41))/(0.4),t=(18+2sqrt(41))/(0.4)

Find the zeros of 200=t(18-0.2t)

The zeros of 200=t(18-0.2t) are t=(18-2sqrt(41))/(0.4),t=(18+2sqrt(41))/(0.4)

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200 = t (18 - 0.2 t)

t = \frac{18 - 2 41}{0.4}, t = \frac{18 + 2 41}{0.4}

Decimal Notation

t = 12.98437 \dots, t = 77.01562 \dots

Solution steps

200 = t (18 - 0.2 t)

Expand

t (18 - 0.2 t) : 18 t - 0.2 t^{2}

200 = 18 t - 0.2 t^{2}

Switch sides

18 t - 0.2 t^{2} = 200

Move

200

to the left side

18 t - 0.2 t^{2} - 200 = 0

Write in the standard form

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

- 0.2 t^{2} + 18 t - 200 = 0

Solve with the quadratic formula

t_{1, 2} = \frac{- 18 \pm 1 8 ^{2} - 4 ( - 0.2 ) ( - 200 )}{2 ( - 0.2 )}

1 8^{2} - 4 (- 0.2) (- 200) = 241

t_{1, 2} = \frac{- 18 \pm 2 41}{2 ( - 0.2 )}

Separate the solutions

t_{1} = \frac{- 18 + 2 41}{2 ( - 0.2 )}, t_{2} = \frac{- 18 - 2 41}{2 ( - 0.2 )}

t = \frac{- 18 + 2 41}{2 ( - 0.2 )} : \frac{18 - 2 41}{0.4}

t = \frac{- 18 - 2 41}{2 ( - 0.2 )} : \frac{18 + 2 41}{0.4}

The solutions to the quadratic equation are:

t = \frac{18 - 2 41}{0.4}, t = \frac{18 + 2 41}{0.4}

∣ 3 x - 2 ∣ = 5

2 x^{2} + 15 x + 18

∣ 7 x - 1 ∣ = 10

x^{2} + 4 x + 8

∣ 2 x - 7 ∣ \geq 1

What are the solutions to the equation 200=t(18-0.2t) ?
The solutions to the equation 200=t(18-0.2t) are t=(18-2sqrt(41))/(0.4),t=(18+2sqrt(41))/(0.4)
Find the zeros of 200=t(18-0.2t)
The zeros of 200=t(18-0.2t) are t=(18-2sqrt(41))/(0.4),t=(18+2sqrt(41))/(0.4)