at which value of x does y=(x+1)(x-19)+7 takeon its minimum value?

y=f(x)= x^2-19x+x-19+7= x^2-18x-12

Find the critical value:
f'(x)= 2x-18, set it equal to zero
0= 2x-18
hence, x=9 is the critical value
Find the second derivative:
f"(x)= 2 & f"(9)= 2, since the second derivative is positive, the function has a minimum at x=2.
x=2 is your answer.