How do you factor and solve # x^2-8x=-3#?

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Answer 1 · 2018-03-07T09:38:36

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Put the equation in form

#x^2-8x+3=0#

From here we have two options:

Option a) Use the solutions formula for 2nd degree equations

#x=(-b+-sqrt(b^2-4ac))/(2a)# for a equ in the canonic form

#ax^2+bx+c=0#

On this way: #x=(8+-sqrt(64-12))/2=(8+-sqrt(4·13))/2=4+-sqrt13#

Option b) complete the square

#x^2-8x+3=(x-4)^2-16+3=(x-4)^2-13=(x-4+sqrt13)(x-4-sqrt13)# from here the solutions are #4+-sqrt13#