How do you solve $x^2 + 16x + 64 = 81$ ?

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Answer 1 · 2018-03-21T16:08:55

$x=1, -17$

rearrange to the form

$ax^2+bx+c=0$

then factorise

$x^2+16x+64=81$

$x^2+16x-17=0$

$(x+17)(x-1)=0$

$" either "x+17=0=>x=-17$

$" or "x-1=0=>x=1$

$:.x=1,-17$

How do you solve x^2 + 16x + 64 = 81x^2 + 16x + 64 = 81?