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Have you learned the quadratic formula?

yea but dat aint working it would be a=4 b=-20 c=1 can u try

a = -4
b = 20
c = 1

yea

im trying 2 figure dis out

im stuck on the step with the quadratic formula

So what is the initial velocity?

Plug that in for v. What is the height above the ground? Plug that in for c.

v is 80

And c?

by-bye polpok

c is 4

0=-16^2+80t+4 simplified its 0=-4t^2+20t+1

Yep.

but i need t

Ok so using the formula for the quadratic, what is a, b, and c.

\[0 =at^2 + bt + c \implies t = \frac{-(b) \pm \sqrt{(b)^2 - 4(a)(c)}}{2(a)}\]

a=-4
b=20
c=1

yea but idk y i keep getting really big numbers wat do u get

t has 2 be 5.1s or 1s or 2s,3s

I got -0.049509756796392246, and 5.049509756796392

Are you forgetting to divide by -8?

Start with the part under the square root

so dont put -20 first

20*20 - 4*(-4)*(1)

What do you get for that part

416

Now take the square root of that

20.3

Ok, now you have 2 solutions:
\[x = \frac{-20 \pm 20.3}{2(-4)}\]

Oh, and you didn't round up.. I think it's 20.396 or 20.4

im still getting 50.75

-20 + 20.4 = ?
-20 - 20.4 = ?

2(-4) = ?

.4

-8

ok so u dont multiply the -20 to the 20.4/-8

No, you add and subtract them

(-20 plus or minus 20.4) / -8

i got -0.5

There's one more possibility.

nope i got -17.45

and -22.55

(-20 - 20.4)/-8

-0.05 is one option.

But the other one is positive

ok i got u now wow thanks but its weird how i cant put the whole thing in the calculator

Because the order matters

ok can u help me wit another one

probably

So how would you set this one up?

i got 100t=-16t^2+80t+4

it's just 100, h is the height.

so \[100 = -16t^2 + 80t + 4\]

ok wat next

move the 100 to the other side so it equals 0 and then use the quadratic formula again.

by subtracting 100 from both sides that is

0=-4t^2+20+24

4-100 is -196, not positive

i got 3

err -96 rather

well den dis aint working now idk u do it

You did it right, I don't know how, but 3 is one answer. There is another answer

If you used -4t^2+20+24
You should have gotten (-1 and 6) though

But the correct formula
-4t^2+20t-24 = 0 gives one of the solutions being 3 and the other being 2

the answer can only be 1s or 2s,3s

show me ur formula

\[0=-4t^2 + 20t - 24 \implies t = \frac{-20 \pm \sqrt{20^2 - 4(-4)(-24)}}{2(-4)}\]

ok i got dat but now getting big numbers again

Do the part under the square root first

\(20^2 - 4(-4)(-24)\)

16

and the square root of 16 is?

Ok, and what's the denominator? 2(-4)

-8

polpak, i have some confusion. check your equation once again

So you have two solutions:
(-20 + 4)/-8 and (-20 - 4)/-8

for the first one you have -16/-8 and the second is -24/-8

omg thanks

Hrm India?

oh whoopse

No.. wait, that's right

ok linda dis is my question create ur own i need help 2 get off mines

ah. ok

can u check the answer i get

i cant do \[\sqrt{-21}\]

so how do i do dis problem

it is a imaginary number!!!! it will come \[i \sqrt{21}\]

does dat equal -1

or 1

\[-1 = i ^{2}\]

i=1 right

well it has 2 be possible cuz there is no i answer on dis paper

one thing, time can't be negative.

your solution seems wrong!!!

i have 0=-4t^+5-1

\[-4t ^{5}-1\]

do you want to write this equation???

my bad 0=-4^2+5-1

i know the answer has 2 be 1 or 1.65 but idk y it aint working

still something wrong in the equation. there is no "t"

0=-4t^2+5t-1

yes, the equation is giving t = 1 and o.5

how uget dat did u use the quadratic formula

hmm, yes

how cuz i got -41/-8 + or - 5

you must be wrong cuz i used calculator to solve.

i did 2 show me wat u put in

\[\frac{-b \pm\sqrt{b^2-4ac}}{2a}\]

i know dat wat numbers did u end up wit

1 and 0.5

will be the answer

no i mean like wat was b^2-4ac

0=-4t^2+5t-1
put a = -4 b = 5 and c = -1

i did i keep getting -41 on top

\[b ^{2}-4ac = 5^{2}-4(-4)(-1)\]

you will get 9

i was putting -5^2

-5^2 will also be 25. the sign doesn't matter when there is a square

thanks

my pleasure!!!

0=-16t^2+27+1 need some help