Simple question 2?

Question

That makes total sense thx heaps. I tried to do the next one with same formula.

S=6, u=11, a=4
S=ut + 1/2 at (squared)

6=11t + (1/2) 4t (squared)
0=11t + 2t (squared) - 6
0=2t (squared) + 11t - 6
0=t (squared) + 11t - 12
0= (t-1)(t+12)
t=1, -12 therefore 1...

But answer is 1/2, -6?? Help again please?? Im trying..

Answer 1

I don't have the question but...

In the end you will find a equation that is in the "form" ax²+bx+c=0, solve it with this formula:

-b+-√(b²-4ac)/2a

I didn't learn math on the english way but i'm just this way is pretty easy lol

Answer 2

6=11t + (1/2) 4t (squared)
0=11t + 2t (squared) - 6
0=2t (squared) + 11t - 6
0=t (squared) + 11t - 12 <--You went wrong there
0= (t-1)(t+12)
t=1, -12 therefore 1...


At that point it should be t²+5 1/2t-3=0
However, I would find it easier to factorise from 2t²+11t-6=0, giving (2t-1)(t+6)=0, and so t=1/2, -6

Answer 3

s = u t + (1 / 2) a t ²
6 = 11 t + (1 / 2) (4) t ²
6 = 11 t + 2 t ²
2 t ² + 11 t - 6 = 0
( t + 6 )( 2t - 1 ) = 0
t = 1 / 2 , t = - 6