Question

Find the real and complex solution of the equation $2x^2-3x+2=0$.

Short answer

The complex solutions are $x=\frac{3+i\sqrt{7}}{4},\frac{3-i\sqrt{7}}{4}$.

Step-by-step solution

Step 1. Apply quadratic formula.

$\begin{array}{c}2x^2-3x+2=\mathrm{0....}\left(\text{Given}\right)\\ \text{or,}x=\frac{-\left(-3\right)\pm \sqrt{{\left(-3\right)}^2-4\left(2\right)\left(2\right)}}{2\left(2\right)}\\ x=\frac{3\pm \sqrt{-7}}{4}\mathrm{....}\left(1\right)\end{array}$

Step 2. By concept of complex numbers.

Further Solving for (1),

$\begin{array}{c}x=\frac{3\pm i\sqrt{7}}{4}\mathrm{....}\left(\sqrt{-1}=i\right)\\ \text{or,}x=\frac{3+i\sqrt{7}}{4},\frac{3-i\sqrt{7}}{4}\end{array}$

Step 3. Description of steps.

Since the roots are complex hence it has complex solution $x=\frac{3+i\sqrt{7}}{4},\frac{3-i\sqrt{7}}{4}$.