Part A.
The quadratic equation,
[tex]ax^2+bx+c=0[/tex]
is equivalent to
[tex]a(x+\frac{b}{2a})^2=\frac{b^2}{4a}-c[/tex]
In our case a=1, b=-3 and c=-18. Then, by substituting these value into the last result, we have
[tex](x+\frac{-3}{2(1)})^2=(\frac{-3}{2(1)})^2+18[/tex]
which gives
[tex]\begin{gathered} (x-\frac{3}{2})^2=\frac{9}{4}+18 \\ (x-\frac{3}{2})^2=\frac{9}{4}+18 \\ (x-\frac{3}{2})^2=\frac{9+72}{4} \\ (x-\frac{3}{2})^2=\frac{81}{4} \end{gathered}[/tex]
Therefore, the answer for part A is:
[tex](x-\frac{3}{2})^2=\frac{81}{4}[/tex]
Part B.
Now, we need to solve the last result for x. Then, by applying square root to both sides, we have
[tex]x-\frac{3}{2}=\pm\sqrt[]{\frac{81}{4}}[/tex]
which gives
[tex]x-\frac{3}{2}=\pm\frac{9}{2}[/tex]
then, by adding 3/2 to both sides, we obtain
[tex]x=\frac{3}{2}\pm\frac{9}{2}[/tex]
Then, we have 2 solutions,
[tex]\begin{gathered} x=\frac{3}{2}+\frac{9}{2}=\frac{12}{2}=6 \\ \text{and} \\ x=\frac{3}{2}-\frac{9}{2}=\frac{-6}{2}=-3 \end{gathered}[/tex]
Therefore, the answer for part B is: -3, 6
