# How do you solve 2x²-(5/2)x+2=0?

Jul 2, 2017

$x = \frac{5}{8} + i \frac{\sqrt{39}}{8}$ or $\frac{5}{8} - i \frac{\sqrt{39}}{8}$

#### Explanation:

$2 {x}^{2} - \frac{5}{2} x + 2 = 0$ can be written as

$2 \left({x}^{2} - \frac{5}{4} x + 1\right) = 0$ - dividing by $2$

or ${x}^{2} - \frac{5}{4} x + 1 = 0$

or ${x}^{2} - 2 \times \frac{5}{8} \times x + {\left(\frac{5}{8}\right)}^{2} - {\left(\frac{5}{8}\right)}^{2} + 1 = 0$

or ${\left(x - \frac{5}{8}\right)}^{2} - \frac{25}{64} + 1 = 0$

or ${\left(x - \frac{5}{8}\right)}^{2} + \frac{39}{64} = 0$

or ${\left(x - \frac{5}{8}\right)}^{2} - \left(- 1 \times \frac{39}{64}\right) = 0$

or ${\left(x - \frac{5}{8}\right)}^{2} - {\left(i \frac{\sqrt{39}}{8}\right)}^{2} = 0$

or $\left(x - \frac{5}{8} + i \frac{\sqrt{39}}{8}\right) \left(x - \frac{5}{8} - i \frac{\sqrt{39}}{8}\right) = 0$

Hence $x = \frac{5}{8} + i \frac{\sqrt{39}}{8}$ or $\frac{5}{8} - i \frac{\sqrt{39}}{8}$