Solution:
Given a quadratic equation;
[tex]ax^2+bx+c=0[/tex]
The roots x of the equation is;
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Given;
[tex]x^2-2x-12=0[/tex][tex]a=1,b=-2,c=-12[/tex]
The values of x are;
[tex]\begin{gathered} x=\frac{-(-2)\pm\sqrt[]{(-2)^2-4(1)(-12)}}{2(1)} \\ x=\frac{2\pm\sqrt[]{4+48}}{2(1)} \\ x=\frac{2\pm\sqrt[]{52}}{2} \\ x=\frac{2\pm2\sqrt[]{13}}{2} \\ x=1\pm\sqrt[]{13} \end{gathered}[/tex]
Then, we can separate the solution;
[tex]x=1+\sqrt[]{13},x=1-\sqrt[]{13}[/tex]
Hence, the roots of the quadratic equation are;
[tex]\begin{gathered} x=1+\sqrt[]{13} \\ x=1-\sqrt[]{13} \end{gathered}[/tex]
