Recall that the quadratic formula states that the solutions to the equation:
[tex]ax^2+bx+c=0[/tex]are:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.[/tex]Therefore the solutions to the given equation are:
[tex]x=\frac{-(-1)\pm\sqrt{(-1)^2-4(6)(-12)}}{2(6)}.[/tex]Simplifying the above result we get:
[tex]x=\frac{1\pm\sqrt{1+288}}{12}=\frac{1\pm\sqrt{289}}{12}=\frac{1\pm14}{12}.[/tex]Then the solutions to the given equation are:
[tex]x=\frac{1-14}{12}=-\frac{13}{12}\text{ and x=}\frac{1+14}{12}=\frac{15}{12}=\frac{5}{4}.[/tex]Answer:
[tex]x=-\frac{13}{12}\text{ and }x=\frac{5}{4}.[/tex]