This technique is valid only when the coefficient of x2 is 1.
1) Transpose the constant term to the right
x2 + 6x = −2.
<span> 2) <span>Add a square number to both sides -- add the square of half the coefficient of x. In this case, add the square of half of 6; that is, add the square of 3, which is 9:</span></span>
x2 + 6x + 9 = −2 + 9.
The left-hand side is now the perfect square of (x + 3).
(x + 3)2 = 7.
3 is half of the coefficient 6.
That equation has the form
<span><span><span>a2</span> = b</span> which implies<span>a = <span>±.</span></span> Therefore,<span><span>x + 3</span> = ±</span> <span>x = <span>−3 ±.</span></span></span>
That is, the solutions to
x2 + 6x + 2 = 0