Answer :
We are given the following equation:
[tex]x^{2}-4x+85=0[/tex]To solve for "x" we will use the quadratic formula, which is:
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]Where the values of "a", "b" and "c" are the coefficients of the equation:
[tex]ax^2+bx+c[/tex]Therefore, in this case, we have:
[tex]\begin{gathered} a=1 \\ b=-4 \\ c=85 \end{gathered}[/tex]Replacing in the quadratic formula:
[tex]x=\frac{-(-4)\pm\sqrt[]{(-4)^2-4(1)(85)}}{2(1)}[/tex]Solving the operations inside the radical:
[tex]x=\frac{-(-4)\pm\sqrt[]{-324}}{2}[/tex]Now we divide the number inside the radical a the product of -1 and 324:
[tex]x=\frac{-(-4)\pm\sqrt[]{(-1)(324)}}{2}[/tex]Now we divide the square root:
[tex]x=\frac{-(-4)\pm\sqrt[]{(-1)}\sqrt[]{324}}{2}[/tex]The square root of -1 is the imaginary unit "i". Therefore, solving the square roots we get:
[tex]x=\frac{-(-4)\pm18i}{2}[/tex]Now we separate the fraction:
[tex]x=\frac{-(-4)}{2}\pm\frac{18i}{2}[/tex]Solving the operations:
[tex]x=2\pm9i[/tex]Therefore, the solutions of the equation are:
[tex]\begin{gathered} x_1=2+9i \\ x_2=2-9i \end{gathered}[/tex]