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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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:

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

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