# How do you solve 2x^2+6x-3=0 by completing the square?

Apr 29, 2015

In this way:

$2 {x}^{2} + 6 x - 3 = 0 \Rightarrow 2 \left({x}^{2} + 3 x\right) - 3 = 0 \Rightarrow$

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

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

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

$2 {\left(x + \frac{3}{2}\right)}^{2} = \frac{9 + 6}{2} \Rightarrow {\left(x + \frac{3}{2}\right)}^{2} = \frac{15}{4} \Rightarrow$

$x + \frac{3}{2} = \pm \frac{\sqrt{15}}{2} \Rightarrow x = - \frac{3}{2} \pm \frac{\sqrt{15}}{2}$.