Answer:
[tex]\displaystyle x_1=-1+\sqrt{3}i[/tex]
[tex]\displaystyle x_2=-1-\sqrt{3}i[/tex]
Step-by-step explanation:
Second-Degree Equation
The second-degree equation or quadratic equation has the general form
[tex]ax^2+bx+c=0[/tex]
where a is non-zero.
There are many methods to solve the equation, one of the most-used is by using the solver formula:
[tex]\displaystyle x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The equation of the question has the values: a=1, b=2, c=4, thus the values of x are
[tex]\displaystyle x=\frac{-2\pm \sqrt{2^2-4\cdot 1\cdot 4}}{2\cdot 1}[/tex]
[tex]\displaystyle x=\frac{-2\pm \sqrt{-12}}{2}[/tex]
Since the square root has a negative argument, both solutions for x are imaginary or complex. Simplifying the radical
[tex]\displaystyle x=\frac{-2\pm 2\sqrt{-3}}{2}=-1\pm\sqrt{3}i[/tex]
The solutions are
[tex]\displaystyle x_1=-1+\sqrt{3}i[/tex]
[tex]\displaystyle x_2=-1-\sqrt{3}i[/tex]
