Question

Find all real and complex solutions of the equation$2x^2+4x+3=0$

Short answer

The solutions is$-1+\frac{1}{\sqrt{2}}i\mathrm{and}-1-\frac{1}{\sqrt{2}}i$

Step-by-step solution

Step 1. Definition of quadratic equations.

A quadratic equation is an equation of the form

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

Where $a,b$and $c$are real numbers with $a\ne 0$

Step 2. Concept of real and complex solution.

If$a\ne 0$, then the solutions of the quadratic equation $a{x}^2+bx+c=0$are,$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

If $b^2-4ac<0$, then the equation has no real solution, it has complex solution and if$b^2-4ac\ge 0$ then equation has real solution

Step 3. Determining the solution.

By the quadratic formula, we have

$\begin{array}{c}x=\frac{-4\pm \sqrt{16-24}}{4}\\ =\frac{-4\pm \sqrt{-8}}{4}\\ =\frac{-4\pm 2\sqrt{2}i}{4}\\ =\frac{-2\pm \sqrt{2}i}{2}\\ =-1+\frac{1}{\sqrt{2}}i\end{array}$

So, the solutions are$-1+\frac{1}{\sqrt{2}}i\mathrm{and}-1-\frac{1}{\sqrt{2}}i$