# How do you solve the equation 3x^2+6x=2x^2+3x-4 by completing the square?

Jul 10, 2017

$x = - \frac{3}{2} - \frac{\sqrt{7}}{2} \setminus i , - \frac{3}{2} + \frac{\sqrt{7}}{2} \setminus i$

#### Explanation:

We have:

$3 {x}^{2} + 6 x = 2 {x}^{2} + 3 x - 4$

First let us collect terms on the LHS

$3 {x}^{2} - 2 {x}^{2} + 6 x - 3 x - 4 = 0$
$\therefore {x}^{2} + 3 x + 4 = 0$

Now we complete the square, by forming a term with half the "$b$" value; that is

$\therefore {\left(x + \frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} + 4 = 0$
$\therefore {\left(x + \frac{3}{2}\right)}^{2} - \frac{9}{4} + 4 = 0$

Which we can now re-arrange and solve;

$\therefore {\left(x + \frac{3}{2}\right)}^{2} = - \frac{7}{4}$

$\therefore x + \frac{3}{2} = \pm \frac{\sqrt{7}}{2} \setminus i$

$\therefore x = - \frac{3}{2} \pm \frac{\sqrt{7}}{2} \setminus i$

Leading to the two complex conjugate solutions:

$x = - \frac{3}{2} - \frac{\sqrt{7}}{2} \setminus i , - \frac{3}{2} + \frac{\sqrt{7}}{2} \setminus i$