Trigonometry Question?

Question

How do I Solve:

2t^2 - 3 = t

(If there is more than one solution, separate them with commas.)

Thanks!

Answer 1

2t^2 -- 3 = t
Or 2t^2 -- t -- 3 = 0
whence
t = 1/4 ± 1/4√{(--1)^2 -- 4(2)(--3)} = 1/4{1 ± 5} = 3/2, -- 1

Answer 2

it is a quadratic equation therefore move everything to one side.

therefore:
2t^2 - t - 3 = 0

factorize(2t - 3)(t + 1) = 0
ab = 0 if and only is a is 0 or b is 0

therefore,

2t - 3 = 0 or t + 1 = 0
t = 3/2 or t = -1

hope that helps,

Kempos

Answer 3

2t^2 - 3 = t

bringing t to the left side,

2t^2 - t - 3 = 0

we now proceed to factorise the expression on the left side.
2t^2 + 2t - 3t - 3 = 0

2t( t + 1) -3( t + 1) = 0

since (t +1) is common to both terms,

(2t - 3)(t + 1) = 0

2t - 3 = 0

t = 3/2

or t + 1= 0

t = -1

the two roots are 3/2 and 1.

Answer 4

2t^2 - 3 = t
2t^2 - t -3=0

ab= product = 2* -3= -6
a+b= sum = -1
a=2 b=-3

2t^2 +2t -3t - 3 =0

2t(t+1) - 3(t+1)=0

(2t-3) (t+1) =0

Therefore,
either (2t-3) = 0 or (t+1) = 0

i.e, t = 3/2 or t = -1

Answer 5

It is not a trigonometric question
you can write 2t^2-3-t=0
the solutions are
t1 = 1+(1+24)^0.5/4 = 1+5/2= 1.5
and t2 = 1-(1+24)^0.5/4 = 1+5/2= -1

roots -1, +1.5

Answer 6

2t^2 - 3 = t ²
2t ² -t - 3 = 0
(2t -3)(t +1) = 0
t = 3/2 . . . and . . . t = - 1

Answer 7

2t^2-3=t

=> 2t^2-t-3 = 0 (which is a quadratic equation of the format ax^2+bx+c=0)

the solution is x = (-b+(sqrt(b^2-4ac)))/2a and (-b-(sqrt(b^2-4ac)))/2a

which is => t= (1+sqrt(1+24))/4 and (1-sqrt(1+24))/4

implies => t = 3/2 and -1