How do you solve #x^2-5x+6=0 #?

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Answer 1 · 2016-04-19T05:44:41

The solutions are
#color(blue)( x= 3#

#color(blue)( x = 2#

#x^2 - 5x + 6 = 0#

The equation is of the form #color(blue)(ax^2+bx+c=0# where:
#a=1, b=-5, c= 6#

The Discriminant is given by:

#Delta=b^2-4*a*c#

# = (-5)^2-(4* 1 * 6)#

# = 25 - 24 = 1 #

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

#x = (-(-5)+-sqrt(1))/(2*1) = (5 +- 1 )/2#

#color(blue)( x = ( 5 +1)/2 = 6/2 = 3#

#color(blue)( x = ( 5 -1)/2 = 4 /2 = 2#