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

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Answer 1 · 2016-06-12T12:09:20

#x=3# or #x=2#

#x^2+6 = 5*x #

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

It can be written as two factors in the form:

#(x+a)*(x+b) =0#.

Where a, b are two integers

such that :
# a + b = -5#

and

# a* b = 6#

Thus, # a=-2 # and #b= -3#

Substituting the values of a and b we have:

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

#(x-3)*(x-2)=0#

#x-3=0#
Or
#x-2=0#

so,
#x=3#
or
#x=2#