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

Mar 3, 2018

color(brown)(x = +-7/4 - 3/4 = 1, -5/2

#### Explanation:

$2 {x}^{2} + 3 x - 5 = 0$

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

(x^2 + (2 * (3/4) * x ) = 5/2

Add ${\left(\frac{3}{4}\right)}^{2}$ to both sides.

${x}^{2} + \left(2 \cdot \left(\frac{3}{4}\right) \cdot x\right) + {\left(\frac{3}{4}\right)}^{2} = \frac{5}{2} + {\left(\frac{3}{4}\right)}^{2} = \frac{49}{16}$

${\left(x + \frac{3}{4}\right)}^{2} = {\left(\sqrt{\frac{49}{16}}\right)}^{2} = {\left(\frac{7}{4}\right)}^{2}$

$x + \frac{3}{4} = \pm \frac{7}{4}$

color(brown)(x = +-7/4 - 3/4 = 1, -5/2