How do you find the roots for #x^2 + 2x – 48 = 0 #?

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Answer 1 · 2015-05-16T07:44:53

#0 = x^2+2x-48#

#= (x+1)^2 - 1 - 48 = (x+1)^2-49 = (x+1)^2 - 7^2#

So #(x+1)^2 = 7#

So #x+1=+-sqrt(7)#

So #x=-1+-sqrt(7)#

In general

#ax^2+bx+c = a(x+b/(2a))^2 + (c - b^2/(4a))#