# How do you solve using the completing the square method x^2 - 8x -48 = 0?

Feb 28, 2016

${x}^{2} - 8 x - 48 = 0$
$\implies {x}^{2} - 2 \cdot x \cdot 4 + {4}^{2} - 16 - 48 = 0$
$\implies {x}^{2} - 2 \cdot x \cdot 4 + {4}^{2} = 64$
$\implies {\left(x - 4\right)}^{2} = 64$
$\implies \left(x - 4\right) = \pm \sqrt{64} = \pm 8$
$\therefore x = 8 + 4 = 12$ and $x = - 8 + 4 = - 4$