Math is hard

let r and s be the root of the quadratic x^2 + bx + c where b and c are constants. If (r-1)(s-1)=7, find b+c

the roots of  1x^2+bx+c     are when    

 x^2 + bx + c  = 0

\(x = {-b \pm \sqrt{b^2-4*1*c} \over 2*1}\\ x = {-b \pm \sqrt{b^2-4c} \over 2}\\ r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\ ~\\ r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\~\\ (r-1)(s-1)=7\\ rs-s-r+1=7\\ rs-(s+r)=6\\~\\ rs=\frac{b^2-(b^2-4c)}{4}=c\\~\\ s+r=-b\\~\\ rs-(s+r)=6\\ c--b=6\\ c+b=6 \)

You should check this answer.

LaTex

x = {-b \pm \sqrt{b^2-4*1*c} \over 2*1}\\
x = {-b \pm \sqrt{b^2-4c} \over 2}\\
r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\
~\\
r= {-b + \sqrt{b^2-4c} \over 2}\qquad s= {-b - \sqrt{b^2-4c} \over 2}\\~\\
(r-1)(s-1)=7\\
rs-s-r+1=7\\
rs-(s+r)=6\\~\\
rs=\frac{b^2-(b^2-4c)}{4}=c\\~\\
s+r=-b\\~\\
rs-(s+r)=6\\
c--b=6\\
c+b=6