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How do you solve #x^2 – 6x -7 = 0#?

1 Answer

Konstantinos Michailidis

Mar 16, 2016

We can write this as follows

#x^2-6x-6-1=(x^2-1)-6(x+1)=(x-1)(x+1)-6(x+1)= (x+1)*(x-1-6)=(x+1)*(x-7)#

Now from #x^2-6x-7=0=>(x+1)*(x-7)=0# we get that the roots are

#x=-1#, #x=7#

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