# How do you solve 2x^2-3x=-3 using the quadratic formula?

Mar 18, 2017

The solutions are $S = \left\{\frac{3}{4} + \frac{\sqrt{15}}{4} i , \frac{3}{4} - \frac{\sqrt{15}}{4} i\right\}$

#### Explanation:

$a {x}^{2} + b x + c = 0$

Here, we have

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

The discriminant is

$\Delta = {b}^{2} - 4 a c = 9 - 4 \cdot 2 \cdot 3 = 9 - 24 = - 15$

As $\Delta < 0$, there are no real roots.

The roots are imaginary

$x = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$x = \frac{3 \pm i \sqrt{15}}{4}$

${x}_{1} = \frac{3}{4} + \frac{\sqrt{15}}{4} i$

${x}_{2} = \frac{3}{4} - \frac{\sqrt{15}}{4} i$