How do you solve $x^2 + 6x = 3$ using completing the square?

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Answer 1 · 2015-07-23T02:36:00

To complete the square you add $9$ to both sides.

Method : you halve the $x$ -coefficient ( $6$ ) to get $3$ , and the square of that ( $9$ ) is needed to complete the square.

$->x^2+6x+9=3+9$

$->(x+3)^2=12->x+3=+-sqrt12=+-2sqrt3$

$->x_(1,2)=-3+-2sqrt3$

(remember $+-$ means two solutions, one with + and one with -)

How do you solve x^2 + 6x = 3x^2 + 6x = 3 using completing the square?