im having a problem with these quadratics I will include a picture

ANSWER and EXPLANATION
a) We want to solve the quadratic equation by completing the square:
[tex]r^2+16r-48=-162[/tex]The first step is to add 48 to both sides of the equation to eliminate -48 from the left-hand side:
[tex]\begin{gathered} r^2+16r-48+48=-162+48_{}^{}_{} \\ r^2+16r=-114 \end{gathered}[/tex]Now, to complete the square, divide 16 by 2 and find the square. Then, add that to both sides of the equation:
[tex]\begin{gathered} r^2+16r+(\frac{16}{2})^2=-114+(\frac{16}{2})^2 \\ r^2+16r+64=-114+64 \\ \Rightarrow(r+8)^2=-50 \end{gathered}[/tex]That is the equation after completing the square.
To find the solutions of r, find the square root of both sides of the equation and simplify:
[tex]\begin{gathered} r+8=\sqrt[]{-50} \\ r+8=\pm5\sqrt[]{2}i \\ \Rightarrow r=-8+5\sqrt[]{2}i;r=-8-5\sqrt[]{2}i \end{gathered}[/tex]Those are the solutions.
b) We want to solve the quadratic equation given by completing the square:
[tex]m^2+4m-20=44[/tex]The first step is to add 20 to both sides of the equation to eliminate -20 from the left-hand side:
[tex]\begin{gathered} m^2+4m-20+20=44+20 \\ \Rightarrow m^2+4m=64 \end{gathered}[/tex]Now, to complete the square, divide 4 by 2 and find the square. Then, add that to both sides of the equation:
[tex]\begin{gathered} m^2+4m+(\frac{4}{2})^2=64+(\frac{4}{2})^2 \\ m^2+4m+4=64+4 \\ (m+2)^2=68 \end{gathered}[/tex]That is the equation after completing the square.
To find the solutions of m, find the square root of both sides of the equation and simplify:
[tex]\begin{gathered} m+2=\sqrt[]{68} \\ m+2=\pm2\sqrt[]{17} \\ \Rightarrow m=-2+2\sqrt[]{17};m=-2-2\sqrt[]{17} \end{gathered}[/tex]Those are the solutions.