# How do you solve by completing the square: 7w² + 6w + 7 = 0?

Mar 30, 2015

In this way:

$7 {w}^{2} + 6 w + 7 = 0 \Rightarrow 7 \left({w}^{2} + \frac{6}{7} w\right) + 7 = 0 \Rightarrow$

$7 \left({w}^{2} + \frac{6}{7} w + {\left(\frac{3}{7}\right)}^{2} - {\left(\frac{3}{7}\right)}^{2}\right) + 7 = 0 \Rightarrow$

$7 \left({w}^{2} + \frac{6}{7} w + {\left(\frac{3}{7}\right)}^{2}\right) - 7 \cdot \frac{9}{49} + 7 = 0 \Rightarrow$

$7 {\left(w + \frac{3}{7}\right)}^{2} - \frac{9}{7} + 7 = 0 \Rightarrow$

$7 {\left(w + \frac{3}{7}\right)}^{2} = \frac{9}{7} - 7 \Rightarrow$

$7 {\left(w + \frac{3}{7}\right)}^{2} = \frac{9 - 63}{7} \Rightarrow$

$7 {\left(w + \frac{3}{7}\right)}^{2} = - \frac{54}{7}$

that is an impossible equation, because a square can't be negative!