2t^2+11-13=0?

2t^2+11-13=0
=>2t^2-2=0
=>2t^2=2
=>t^2=1
=>t= +/-1<===solution

look at 2 and 13. product is 26. are there factors of 26 that will subtract (because of the minus in front of 13) to make that middle 11? (26-1 = 25, 13-2 = 11) yes, (coffee kicking in, thought we'd have to use quad formula). so

(2t + 13)(t - 1) = 0
2t + 13 = 0
t = -13/2
or
t - 1 = 0
t = 1

(2t+13)(t-1) = 0
2t+13 = 0 --> t = -6.5
t-1 =0 --> t = 1

2t² + 11 - 13 = 0
(2t + 13) (t - 1) = 0
2t + 13 = 0 and t - 1 = 0
2t = -13 and t = 1
t = -13/2, 1

2t^2 - 11t - 13 = 0
2t^2 + 2t - 13t - 13 = 0
2t(t + 1) - 13(t + 1) = 0
(t + 1)(2t - 13) = 0

t + 1 = 0
t = -1

2t - 13 = 0
2t = 13
t = 13/2 (6.5)

∴ t = -1 , 13/2 (6.5)

Factor it:

(2t + 13 )(t -1) = 0

Set each factor = to 0:

2t + 13 = 0 or t-1 = 0


2t = -13 or t = 1

t = -13/2 or t = 1