# What kind of equation is this?

1. May 6, 2007

### light_bulb

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.

2. May 6, 2007

3. May 6, 2007

### light_bulb

can you show me how to solve this? i'm getting stuck.

4. May 6, 2007

### d_leet

What have you tried to do already? Have you tried factoring it, or tried applying the quadratic formula?

5. May 6, 2007

### light_bulb

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

Last edited: May 6, 2007
6. May 6, 2007

### cristo

Staff Emeritus
Your last equation is correct, however you have made an arithmetic error-- it does not give the solutions -2 and 18.

7. May 6, 2007

### d_leet

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

[20 ± sqrt{400 - 4(1)(96)}]/2(1) = t

8. May 6, 2007

### light_bulb

your both right, {-8,12} but that still doesn't seem right?

9. May 6, 2007

### sstone

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.

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

$$\frac { 20 \pm 28 } {2}$$

Last edited: May 6, 2007
10. May 6, 2007

### light_bulb

-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. May 6, 2007

### Office_Shredder

Staff Emeritus
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. May 6, 2007

thanks