How do you solve #y^2-3y+2=0#?

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Answer 1 · 2016-04-24T07:28:03

The solutions are:
#y = 2#

#y = 1#

#y^2 - 3y + 2 = 0 #

The equation is of the form #color(blue)(ay^2+by+c=0# where:

#a=1, b=-3, c=2#

The Discriminant is given by:
#Delta=b^2-4*a*c#
# = (-3)^2-(4* 1 * 2)#
# = 9 - 8 = 1#

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

#y = (-(-3)+-sqrt(1))/(2*1) = (3 +- 1 )/2#

#y = (3 + 1 ) /2 = 4 / 2 = 2#

#y = (3 - 1 ) /2 = 2 / 2 = 1#