# How do you solve the quadratic with complex numbers given 2x^2-6x+7=0?

Sep 30, 2016

$x = \frac{3}{2} + i \frac{\sqrt{5}}{2}$ or $x = \frac{3}{2} - i \frac{\sqrt{5}}{2}$

#### Explanation:

$2 {x}^{2} - 6 x + 7 = 0$ - we first ty to make complete square as in ${\left(x + a\right)}^{2}$

$2 {x}^{2} - 6 x + 7 = 0$

$\Leftrightarrow {x}^{2} - 3 x + \frac{7}{2} = 0$

or ${x}^{2} - 2 \times \frac{3}{2} \times x + {\left(\frac{3}{2}\right)}^{2} - {\left(\frac{3}{2}\right)}^{2} + \frac{7}{2} = 0$

or ${\left(x - \frac{3}{2}\right)}^{2} - \frac{9}{4} + \frac{14}{4} = 0$

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

or ${\left(x - \frac{3}{2}\right)}^{2} - {\left(\frac{\sqrt{5}}{2} \times i\right)}^{2} = 0$

As this is of the form ${a}^{2} - {b}^{2} = \left(a + b\right) \left(a - b\right)$

The above is equal to

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

Hence $x - \frac{3}{2} - i \frac{\sqrt{5}}{2} = 0$ or $x - \frac{3}{2} + i \frac{\sqrt{5}}{2} = 0$

i.e. $x = \frac{3}{2} + i \frac{\sqrt{5}}{2}$ or $x = \frac{3}{2} - i \frac{\sqrt{5}}{2}$