# How do you solve 4x^2-8=-13x by completing the square?

$y = 4 {x}^{2} + 13 x - 8 = 0.$
$y = {x}^{2} + \frac{13 x}{4} - 2 = 0$
$= {x}^{2} + 13 \frac{x}{4} + \frac{169}{16} - \frac{169}{16} - 2 = 0$
${\left(x + \frac{13}{8}\right)}^{2} = \frac{169}{64} + 2 = \frac{297}{64}$
$x = - \frac{13}{8} + \frac{297}{64} =$
$x = - \frac{13}{8} - \frac{297}{64} =$