# How do you solve by completing the square: -2x^2-3x-3=0?

##### 1 Answer
Mar 31, 2015

$- 2 {x}^{2} - 3 x - 3 = 0$

Divide all terms of both sides of the equation by $\left(- 2\right)$ so we are working with an expression that begins simply with ${x}^{2}$

${x}^{2} + \frac{3}{2} x + \frac{3}{2} = 0$

then move the constant $\left(- \frac{3}{2}\right)$ off the left-side of the equation by subtracting $\frac{3}{2}$ from both sides
${x}^{2} + \frac{3}{2} x = - \frac{3}{2}$

To "complete the square" we need something of the form:
${\left(x + a\right)}^{2}$
which equals ${x}^{2} + 2 a x + {a}^{2}$

Since we have computed the coefficient of $x$ to be $\frac{3}{2}$
$\rightarrow a = \frac{3}{4}$
and ${a}^{2} = \frac{9}{4}$

Add $\frac{9}{4}$ to both sides of the equation
${x}^{2} + \frac{3}{2} x + \frac{9}{4} = \frac{9}{4} - \frac{3}{2}$

Rewrite the left-side as a square and simplify the right-side
${\left(x + \frac{3}{2}\right)}^{2} = \frac{3}{4}$

Take the square root of both sides
$x + \frac{3}{2} = \pm \frac{\sqrt{3}}{2}$

$x = - \frac{3 + \sqrt{3}}{2}$
or
$x = - \frac{3 - \sqrt{3}}{2} = \frac{\sqrt{3} - 3}{2}$