# How do you solve using completing the square method 2x^2-4x-3=0?

Oct 8, 2016

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

$m$ has to make the expression in the parentheses a perfect square. So, $m = 1$.

$2 \left({x}^{2} - 2 x + 1 - 1\right) = 3$

$2 \left({x}^{2} - 2 x + 1\right) - 2 = 3$

$2 {\left(x - 1\right)}^{2} = 5$

${\left(x - 1\right)}^{2} = 2.5$

x - 1= +-sqrt(2.5

$x = 1 \pm \sqrt{2.5}$

Hopefully this helps!