# How do you solve using the completing the square method 3x^2 + 2x + 5 = 0?

##### 1 Answer
Feb 14, 2017

Solution: $x = - \frac{1}{3} + \frac{\sqrt{14}}{3} \cdot i \mathmr{and} x = - \frac{1}{3} - \frac{\sqrt{14}}{3} \cdot i$

#### Explanation:

$3 {x}^{2} + 2 x + 5 = 0 \mathmr{and} 3 \left({x}^{2} + \frac{2}{3} x\right) + 5 = 0 \mathmr{and} 3 \left({x}^{2} + \frac{2}{3} x + {\left(\frac{1}{3}\right)}^{2}\right) - \frac{1}{3} + 5 = 0$ or $3 {\left(x + \frac{1}{3}\right)}^{2} + \frac{14}{3} = 0 \mathmr{and} {\left(x + \frac{1}{3}\right)}^{2} = - \frac{14}{9} \mathmr{and} x + \frac{1}{3} = \pm \sqrt{- \frac{14}{9}}$
or $x = - \frac{1}{3} \pm \sqrt{- \frac{14}{9}} \therefore x = - \frac{1}{3} + \frac{\sqrt{14}}{3} \cdot i \mathmr{and} x = - \frac{1}{3} - \frac{\sqrt{14}}{3} \cdot i$ [i^2 = -1]
Solution: $x = - \frac{1}{3} + \frac{\sqrt{14}}{3} \cdot i \mathmr{and} x = - \frac{1}{3} - \frac{\sqrt{14}}{3} \cdot i$ [Ans]