How do you solve #2t^2 – 6t + 1 = 0# using the quadratic formula?

Question

1521 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-14T06:41:40

The solutions are:
#color(blue)(t=((3+sqrt(7)))/2#

#color(blue)(t=((3-sqrt(7)))/2#

#2t^2-6t+1#

The equation is of the form #color(blue)(at^2+bt+c=0# where:
#a=2, b=-6, c=1#

The Discriminant is given by:

#Delta=b^2-4*a*c#
# = (-6)^2-(4*2*1)#
# = 36- 8=28#

The solutions are found using the formula
#t=(-b+-sqrtDelta)/(2*a)#

#t = (-(-6)+-sqrt(28))/(2*2) = (6+-2sqrt(7))/4#

#(6+-2sqrt(7))/4 = (2(3+-sqrt(7)))/4#

#=(cancel2(3+-sqrt(7)))/cancel4#

#=((3+-sqrt(7)))/2#

The solutions are
#color(blue)(t=((3+sqrt(7)))/2#

#color(blue)(t=((3-sqrt(7)))/2#