My first question asked me to write down the greatest value of 1 / (x^2 +6x +20) and another question that I'm stuck on is solve the simultaneous equation of y = x^2 - 3x +2 and y = 3x -7 (I have no problem of doing that). Hence, at the point (3,2) find the equation of the normal to y = x^2 - 3x + 2.

Hello!
1. I suppose your function is  f(x) = 1/(x^2 + 6x + 20). The expression at the denominator is equal to  x^2 + 6x + 9 + 11 = (x+3)^2 + 11 and is always positive.
The greatest value of f(x) is reached at the same point where the least value of  1/f(x) = (x+3)^2 + 11  is reached, which point is clearly x = -3 (because (x+3)^3 is always non-negative and is zero only at x=-3 ).
The value of f at this point is 1/11. This is the answer.
 
2. I agree with your answer for the system of equations. It is the only point (3, 2). Denote g(x)=x^2-3x+2.
The normal is a straight line which goes through the given point and is perpendicular to the curve at the given point. Also the normal is perpendicular to the tangent line at the same point.
The equation of a tangent line at the point (t,g(t)) is  y = (x-t)g'(t)+g(t). We have t=3, g(t)=2, g'(t)=2t-3, g'(3)=3. The equation is  y=(x-3)*3+2=3x-7. Actually it is the line from our system:)  Its slope is m=3.
The perpendicular line has the slope -1/m=-1/3, and its equation is  y=-1/3 x+b. The point (3,2) satisfies this equation so  2=-1/3*3+b=-1+b, thus b=3.
The answer is  y=-1/3x+3.
Or, to put the answer in ax+by+c = 0 form, the answer could also be x/3 + y -3 = 0