# How do you solve by completing the square 2x^2+7x-1=0?

Apr 1, 2015

$2 {x}^{2} + 7 x - 1 = 0$

Remove the constant from the left-side by adding $1$ to both sides
$2 {x}^{2} + 7 x = 1$

Divide all terms by $2$ to reduce the ${x}^{2}$ coefficient to $1$
${x}^{2} + \frac{7}{2} x = \frac{1}{2}$

If the first 2 terms of ${\left(x + a\right)}^{2}$ evaluate as ${x}^{2} + \frac{7}{2} x$
then $a = \frac{7}{4}$

Add (7/4)^2 to both sides of the equation
${x}^{2} + \frac{7}{2} x + {\left(\frac{7}{4}\right)}^{2} = \frac{1}{2} + \frac{49}{16}$

Rewrite the left-side as a square (and simplify the right side)
${\left(x + \frac{7}{4}\right)}^{2} = \frac{57}{16}$

Take the square root of both sides (remember both plus and minus)
$x + \frac{7}{4} = \pm \frac{\sqrt{57}}{4}$

$x = \frac{- 7 + \sqrt{57}}{4}$
or
$x = \frac{- 7 - \sqrt{57}}{4}$