How do you solve #x^2 - 20x + 100 = 36#?

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Answer 1 · 2018-03-24T17:34:55

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We have, #x^2-20x +100-36=0#

Or, #x^2-20x +64=0#

Carrying out middle-term factorization,we get,

#x^2-16x-4x +64=0#

#x(x-16)-4(x-16)=0#

#(x-16)(x-4)=0#

Thus,the two solutions of #x# which we get are #x=16,4#