How do you solve #x^2+10x=3# by completing the square?

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Answer 1 · 2016-03-27T15:07:59

See explanation...

Add #(10/2)^2 = 5^2 = 25# to both sides to get:

#x^2+10x+25 = 28#

The left hand side is equal to #(x+5)^2# so we have:

#(x+5)^2 = 28 = 2^2*7#

Hence:

#x+5 = +-sqrt(2^2*7) = +-2sqrt(7)#

So:

#x = -5+-2sqrt(7)#

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More Info

In general, we find:

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

From which we can derive the quadratic formula for zeros of #ax^2+bx+c#, namely:

#x = (-b+-sqrt(b^2-4ac))/(2a)#