The only formulas you have to know are:
[tex]\begin{gathered} \sqrt{a\cdot b}=\sqrt{a}\cdot\sqrt{b} \\ i^2=-1\rightarrow i=\sqrt{-1} \end{gathered}[/tex]
When you do not know the root of a number, you have to express its root like a product of its main factors, for example:
[tex]\sqrt{75}=5\sqrt{3}[/tex]
To find these factors, we can divide the original number among other numbers and multiply them, for example:
When we know those factors, we can use the laws of roots to simplify:
[tex]\begin{gathered} 75=5^2\cdot3 \\ \sqrt[]{75}=\sqrt{5^2\cdot3}=5^{\frac{2}{2}}\cdot\sqrt{3}=5\sqrt{3} \end{gathered}[/tex]
With this in mind, we can now solve the exercise:
First term:
[tex]\begin{gathered} \frac{-20\pm\sqrt{75}}{5} \\ \\ \frac{-20\pm5\sqrt{3}}{5}\text{ \lparen Divide each term of the numerator by the denominator\rparen} \\ \\ -4\pm\sqrt[]{3} \end{gathered}[/tex]
Second term:
*Notice that
[tex]\sqrt{-81}=\sqrt{(-1)\cdot(81)}=\sqrt{81}\cdot\sqrt{-1}=\sqrt{81}i=9i[/tex][tex]\begin{gathered} \frac{6\pm\sqrt{-81}}{3} \\ \\ \frac{6\pm9i}{3} \\ \\ 2\pm3i \end{gathered}[/tex]
Third term:
*Notice the followings:
[tex]\sqrt{-28}=\sqrt{28\cdot-1}=\sqrt{28}\cdot\sqrt{-1}=\sqrt{4\cdot7}i=2\sqrt{7}i[/tex]
Finally,
[tex]\begin{gathered} \frac{-4\pm\sqrt{-28}}{8} \\ \\ \frac{-4\pm2\sqrt{7}i}{8} \\ \\ \frac{-4}{8}\pm\frac{2\sqrt{7}}{8}i \\ \\ \frac{-1}{2}\pm\frac{\sqrt{7}}{4}\imaginaryI \end{gathered}[/tex]
