How do you solve #x^ { 2} = 14x - 45#?

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Answer 1 · 2017-05-27T00:20:01

#x =5 or x=9#

#x^2 =14x-45" "larr# It is a quadratic equation. Make it #=0#

#x^2 -14x+45 =0" "larr# factorise

Find factors of 45 which add to #14#
These will be #9 and 5#

#(x-9)(x-5)=0#

Set each factor equal to #0# and solve for #x#

#x-9 =0 " "rarr x =9#

#x-5 =0 " "rarr x =5#