Given:
[tex]x^2+3x-6=0[/tex]
This is of the form
[tex]ax^2+bx+c=0[/tex]
where a=1, b=3 and c=-6.
Using the quadratic formuls.
[tex]x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}[/tex]
Substitute a=1, b=3, and c=-6 in the formula, we get
[tex]x=\frac{-3\pm\sqrt[]{3^2-4(1)(-6)}}{2(1)}[/tex]
[tex]x=\frac{-3\pm\sqrt[]{9+24}}{2}[/tex]
[tex]x=\frac{-3\pm\sqrt[]{33}}{2}[/tex]
Hence the roots of the given equation are
[tex]x=\frac{-3+\sqrt[]{33}}{2},x=\frac{-3-\sqrt[]{33}}{2}[/tex]
Options B and C are correct.
