# How do you complete the square for x^2+6x-7=0?

${\left(x + 3\right)}^{2} = 16$

#### Explanation:

The given equation:

${x}^{2} + 6 x - 7 = 0$

${x}^{2} + 2 \setminus \cdot 3 \setminus \cdot x - 7 = 0$

${x}^{2} + 2 \setminus \cdot 3 \setminus \cdot x + {3}^{2} - {3}^{3} - 7 = 0$

${\left(x + 3\right)}^{2} - 16 = 0$

${\left(x + 3\right)}^{2} = 16$