What kind of equation is this?

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The equation 5t^2 - 100t = 480 is identified as a quadratic equation that requires solving for t. The correct approach involves rearranging the equation to 5t^2 - 100t - 480 = 0 and applying the quadratic formula. After correcting arithmetic errors, the solutions are found to be -4 and 24. The discussion highlights the importance of correctly changing the sign when rearranging equations. Overall, the participants emphasize the need for accuracy in calculations and understanding of quadratic equations.
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5t^2 - 100t = 480

solve for t, book has it as one of the practice problems but doesn't show how to work it out.
 
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It's a quadratic equation.
 
can you show me how to solve this? I'm getting stuck.
 
light_bulb said:
can you show me how to solve this? I'm getting stuck.

What have you tried to do already? Have you tried factoring it, or tried applying the quadratic formula?
 
yea i get {-2,18}. it doesn't look right?

heres what I'm doing:

5t^2 - 100t = 480

so

t^2 -20 + 96 = 0

then

[± 20 sqrt{400 - 4(1)(96)}] / 2(1) = t
 
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Your last equation is correct, however you have made an arithmetic error-- it does not give the solutions -2 and 18.
 
light_bulb said:
[sqrt{400 - 4(1)(96)}] ± 20 / 2(1) = t

The order in the numerator is not quite right it should be


[20 ± sqrt{400 - 4(1)(96)}]/2(1) = t
 
your both right, {-8,12} but that still doesn't seem right?
 
Well you forgot to add change 480 to -480 since you flipped sides.
So...5t^2-100t-480=0 now the solutions are -4 and 24 and when you put them in the equation you get the correct solution.

\frac { 20 \pm \sqrt { (-20)^2-4 \cdot 1 \cdot (-96) } } {2 \cdot 1}

\frac { 20 \pm 28 } {2}
 
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  • #10
-4 works :)now if only the book would have mentioned that i needed to change the sign :/

one more time to make sure i got it right.

5t^2 - 100t = 480

so

t^2 -20 - 96 = 0

then

[± 20 sqrt{400 - 4(1)(-96)}] / 2(1) = t

now the calc buttons 4*-96 = 384
400-(-384) = 784
sqrt786 = 28
20-28 = -8
-8/2 = -4

thanks to all that helped!
 
  • #11
The book shouldn't need to mention you change the sign, you should know how subtraction works by the time you deal with quadratic equations

Keep in mind that it's 20±28, so you have a second solution also
 
  • #12
thanks:rolleyes:
 

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