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

Feb 3, 2017

The solutions are $S = \left\{1 , - \frac{5}{2}\right\}$

Explanation:

Let's complete the square

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

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

$2 \left({x}^{2} + \frac{3}{2} x + \frac{9}{16}\right) - 5 - \frac{9}{8} = 0$

$2 {\left(x + \frac{3}{4}\right)}^{2} - \frac{49}{8} = 0$

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

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

Taking the square roots

$\left(x + \frac{3}{4}\right) = \pm \frac{7}{4}$

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

${x}_{1} = - \frac{10}{4} = - \frac{5}{2}$

${x}_{2} = \frac{4}{4} = 1$