# How do you solve using the completing the square method 5v^2-21=10v?

$v = 1 + \sqrt{\frac{26}{5}} \mathmr{and} v = 1 - \sqrt{\frac{26}{5}}$
$5 {v}^{2} - 21 = 10 v \implies 5 {v}^{2} - 10 v = 21 \implies {v}^{2} - 2 v = \frac{21}{\textcolor{red}{5}}$
$\therefore {v}^{2} - 2 v \textcolor{red}{+ 1} = \frac{21}{5} \textcolor{red}{+ 1} \implies {\left(v - 1\right)}^{2} = \frac{26}{5} = {\left(\sqrt{\frac{26}{5}}\right)}^{2}$
$\implies v - 1 = \pm \sqrt{\frac{26}{5}} \implies v = 1 \pm \sqrt{\frac{26}{5}}$
$v = 1 + \sqrt{\frac{26}{5}} \mathmr{and} v = 1 - \sqrt{\frac{26}{5}}$