# How do you solve using the completing the square method 2x^2 + 6x -5 =0 ?

##### 1 Answer
Apr 15, 2018

$x = - \frac{3}{2} \pm \frac{1}{2} \sqrt{19}$

#### Explanation:

$\text{using the method of "color(blue)"completing the square}$

• " the coefficient of the "x^2" term must be 1"

$\text{factor out } 2$

$\Rightarrow 2 \left({x}^{2} + 3 x - \frac{5}{2}\right) = 0$

• " add/subtract "(1/2"coefficient of the x-term")^2" to"
${x}^{2} + 3 x$

$\Rightarrow 2 \left({x}^{2} + 2 \left(\frac{3}{2}\right) x \textcolor{red}{+ \frac{9}{4}} \textcolor{red}{- \frac{9}{4}} - \frac{5}{2}\right) = 0$

$\Rightarrow 2 {\left(x + \frac{3}{2}\right)}^{2} + 2 \left(- \frac{9}{4} - \frac{5}{2}\right) = 0$

$\Rightarrow 2 {\left(x + \frac{3}{2}\right)}^{2} - \frac{19}{2} = 0$

$\Rightarrow 2 {\left(x + \frac{3}{2}\right)}^{2} = \frac{19}{2}$

$\Rightarrow {\left(x + \frac{3}{2}\right)}^{2} = \frac{19}{4}$

$\textcolor{b l u e}{\text{take the square root of both sides}}$

$\sqrt{{\left(x + \frac{3}{2}\right)}^{2}} = \pm \sqrt{\frac{19}{4}} \leftarrow \textcolor{b l u e}{\text{note plus or minus}}$

$\Rightarrow x + \frac{3}{2} = \pm \frac{1}{2} \sqrt{19}$

$\Rightarrow x = - \frac{3}{2} \pm \frac{1}{2} \sqrt{19} \leftarrow \textcolor{red}{\text{exact solutions}}$