Solution
Given the quadratic equation
[tex]x^2-2x-38=0[/tex]
we need to find the zeros of the equation
To do that, we use the completing the square method
Step 1. Add 38 to both sides
[tex]\begin{gathered} \Rightarrow x^2-2x-38+38=38 \\ \\ \Rightarrow x^2-2x=38 \end{gathered}[/tex]
Step 2: add the square of half of the coefficient of x to both sides
That is;
[tex]\begin{gathered} \Rightarrow x^2-2x+(\frac{1}{2}\cdot(-2))^2=38+(\frac{1}{2}\cdot(-2))^2 \\ \\ \Rightarrow x^2-2x+1=38+1 \\ \\ \Rightarrow x^2-2x+1=39 \\ \\ \Rightarrow(x-1)^2=39 \end{gathered}[/tex]
Step 3: Simplify the above expression;
[tex]\begin{gathered} \Rightarrow x-1=\pm\sqrt[]{39} \\ \\ \Rightarrow x=1\pm\sqrt[]{39} \end{gathered}[/tex]
