# Math You may also complete the square as alternate method, using as pattern the formula `(a+b)^2 = a^2+2ab+b^2` .

You need to consider the first two terms of the equation such that:

`x^2 + 6x = x^2 + 2*3*x + 3^2 - 3^2`

Notice that you need to equate the mid term `2ab` ...

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You may also complete the square as alternate method, using as pattern the formula `(a+b)^2 = a^2+2ab+b^2` .

You need to consider the first two terms of the equation such that:

`x^2 + 6x = x^2 + 2*3*x + 3^2 - 3^2`

Notice that you need to equate the mid term `2ab`  and `6x`  such that:

`2ab = 6x`

Since you already know `a = x` , then, you may substitute x for a such that:

`2xb = 6x => b = (6x)/(2x) = 3`

Since you need to add the square of 3 to complete the square, you also need to subtract `3^2` , to keep the balance such that:

`x^2 + 6x = (x+3)^2 - 9`

`x^2 + 6x - 7 = (x+3)^2 - 9 - 7`

Since `x^2 + 6x - 7 = 0` , then `(x+3)^2 - 9 - 7 = 0`  such that:

`(x+3)^2 - 9 - 7 = 0 => (x+3)^2=16 => x+3 = +-sqrt16`

`x+3 = +-4 => x_1 = -3+4 => x_1 = 1`

`x_2 = -3-4 => x_2 = -7`

Hence, evaluating the solutions to the given quadratic equation yields `x_1 = 1`  and `x_2 = -7` .

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