Question 955017
(1,10)
{{{10=a(1)^2+b(1)+5}}}
1.{{{a+b=5}}}
.
.
(2,17)
{{{17=a(2)^2+b(2)+5}}}
{{{4a+2b=12}}}
2.{{{2a+b=6}}}
Subtract eq. 1 from eq. 2,
{{{2a+b-a-b=6-5}}}
{{{a=1}}}
Then from eq. 1,
{{{1+b=5}}}
{{{b=4}}}