This technique is valid only when the coefficient of x2 is 1.
1) Transpose the constant term to the right
x2 + 6x = −2.
2) 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: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 forma2 = b which impliesa = ±. Therefore,x + 3 = ± x = −3 ±.That is, the solutions tox2 + 6x + 2 = 0