# How do you solve the equation 3x^2+5x+4=0 by completing the square?

Dec 21, 2016

$x = \frac{1}{6} \left(- 5 \pm \sqrt{23} i\right)$

#### Explanation:

Given:

$f \left(x\right) = 3 {x}^{2} + 5 x + 4$

I would first premultiply by $3 \cdot {2}^{2} = 12$ to avoid having to deal with fractions very much...

$0 = 12 f \left(x\right)$

$\textcolor{w h i t e}{0} = 12 \left(3 {x}^{2} + 5 x + 4\right)$

$\textcolor{w h i t e}{0} = 36 {x}^{2} + 60 x + 48$

$\textcolor{w h i t e}{0} = 36 {x}^{2} + 60 x + 25 + 23$

$\textcolor{w h i t e}{0} = {\left(6 x + 5\right)}^{2} + 23$

$\textcolor{w h i t e}{0} = {\left(6 x + 5\right)}^{2} - {\left(\sqrt{23} i\right)}^{2}$

$\textcolor{w h i t e}{0} = \left(\left(6 x + 5\right) - \sqrt{23} i\right) \left(\left(6 x + 5\right) + \sqrt{23} i\right)$

$\textcolor{w h i t e}{0} = \left(6 x + 5 - \sqrt{23} i\right) \left(6 x + 5 + \sqrt{23} i\right)$

Hence:

$x = \frac{1}{6} \left(- 5 \pm \sqrt{23} i\right)$