A. 9
Explanationto solve for x we can use the quadratic formula
it says
[tex]\begin{gathered} for \\ ax^2+bc+c=0 \\ the\text{ solution for x is} \\ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \end{gathered}[/tex]hence
Step 1
a) let
[tex]\begin{gathered} x^2-3x-54=0\Rightarrow ax^2+bx+c=0 \\ so \\ a=1 \\ b=-3 \\ c=-54 \end{gathered}[/tex]b) now, replace in the formula and evaluate
[tex]\begin{gathered} x=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ x=\frac{-(-3)\pm\sqrt{(-3)^2-4(1)(-54)}}{2(1)} \\ x=\frac{-(-3)\pm\sqrt{9+216}}{2}=\frac{3\pm\sqrt{225}}{2} \\ x=\frac{3\pm15}{2} \end{gathered}[/tex]so
[tex]\begin{gathered} x_1=\frac{3+15}{2}=9 \\ x_2=\frac{3-15}{2}=-6 \end{gathered}[/tex]therefore, the answer is
A. 9
I hope this helps you
