My Math Forum how did they come to the final step of this equation?

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 April 8th, 2017, 07:11 AM #1 Newbie   Joined: Mar 2017 From: leuven Posts: 20 Thanks: 0 how did they come to the final step of this equation? -0.2tÂ²+1.2t+3.2 = 3.75 <-> -0.2tÂ²+1.2t-0.55=0 <-> t=0.5 or t=5.5 how did they get to that last step? I tried a lot of things, but nothing gave me those values. Does anyone have any idea how they went from -0.2tÂ²+1.2t-0.55=0 to t=0.5 or t=5.5? Last edited by skipjack; April 8th, 2017 at 07:32 AM.
April 8th, 2017, 07:27 AM   #2
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$-0.2t^2+1.2t-0.55 = 0$

multiply every term by $-100$ to clear the decimals ...

$20t^2 - 120t + 55 = 0$

factor out $5$ from every term ...

$5(4t^2 - 24t + 11) = 0$

$5(2t - 1)(2t - 11) = 0$

$t=0.5$, $t= 5.5$

note you can also get out your calculator and use the quadratic formula with the original equation ...
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 April 8th, 2017, 07:40 AM #3 Global Moderator   Joined: Dec 2006 Posts: 18,250 Thanks: 1439 The equation is equivalent to -(t - 0.5)(0.2t - 1.1) = 0, or -0.2(t - 0.5)(t - 5.5) = 0 if you prefer, so one of the factors is zero.
April 12th, 2017, 12:32 PM   #4
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Quote:
 Originally Posted by skeeter $-0.2t^2+1.2t-0.55 = 0$ multiply every term by $-100$ to clear the decimals ... $20t^2 - 120t + 55 = 0$ factor out $5$ from every term ... $5(4t^2 - 24t + 11) = 0$ factor the quadratic ... $5(2t - 1)(2t - 11) = 0$ $t=0.5$, $t= 5.5$ note you can also get out your calculator and use the quadratic formula with the original equation ...
OH GOD! i'm quite sure the second step is what they did in this excercise.

i think they used the quadratic formula, which is a formula i know (didn't know its name though good to know that for the future).

I'l definitly take a look at the other way to solve it

Also sorry for the late reply, i moved location and have been slacking off with my math training for a few days. but today i'm on it again.

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