Question 766252
t^2-2√(2)t+1=0
using quadratic formula to solve for t
{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
a=1, b=-2√2, c=1
b^2=(-2√2)^2=8
4ac=4*1*1=4
..
t=[2√2±√(8-4)]/2
t=[2√2±√4)]/2
t=[2√2±2]/2
t=[√2±1]
t=1.41±1
t≈0.41
or
t≈2.41