Solve the equation for t?

(t-1)^2=2t^2-t-29 . multiply out the left side
t^2 - 2t + 1 = 2t^2-t-29 . move everything to the right
0 = t^2 + t - 30 . factor
0 = (t - 5)(t + 6) . set each factor = to 0 and solve

t - 5=0 or t = 5

t + 6 = 0 or t = -6

(t - 1)^2 = 2t^2 - t - 29
(t - 1)(t - 1) = 2t^2 - t - 29
t^2 - 2t + 1 = 2t^2 - t - 29
2t^2 - t^2 - t + 2t - 29 - 1 = 0
t^2 + t - 30 = 0
(t + 6)(t - 5) = 0

t + 6 = 0
t = -6

t - 5 = 0
t = 5

∴ t = -6 , 5

(t-1)^2 = 2t^2 - t - 29
t^2 - 2t +1 = 2t^2 - t - 29
0 = t^2 + t - 30

The two solutions of a quadratic At^2 + Bt + C are
[-B + sqrt(B^2 - 4AC)]/2A and [-B - sqrt(B^2 - 4AC)]/2A

so we get t = [-1 + sqrt(121)]/2 and t = [-1 - sqrt(121)]/2

sqrt(121) = 11,

so t = 5 or t = -6.

1)expand the left side of the equation

t^2 -2t + 1 = 2t^2 - t - 29

2) Simplify by combining similar terms
(Best if transpose everything to the right side)

2t^2 - t^2 - t + 2t -29 -1 = 0
t^2 + t - 30 = 0

3) Factor the equation

(t+5)(t-6)=0

Therefore:

t = -5
t = 6

t^2-2t+1=2t^2-t+29
t^2+t-30=0
(t-5)(t+6)=0
t=5, or t=-6
come on!
you really need to work harder if you are in high school now.

try expanding the (t-1)^2 and then bringing all that to the opposite side of the equation and solving algebraically (i.e. add the t^2 terms, the t terms and the constant terms). The equation will have to equal zero but if a t^2 exists then it is a polynomial and thus has 2 answers.

(t-1)^2 = t^2 -2t +1
so you have t^2-2t+1=2t^2-t-29
subtract t^2 from both side you get
-2t+1=t^2-t-29
add 2t to each side
you get 1= t^2+t-29
0=t^2+t-30
(t+6)(t-5)
t=-6 or t= +5

t ² - 2t + 1 = 2 t ² - t - 29
0 = t ² + t - 30
( t + 6 ) ( t - 5 ) = 0
t = - 6 , t = 5