Question 926793
Given quadratic pattern:
{{{4}}};{{{9}}};{{{x}}};{{{37}}}

quadratic:{{{an^2 + bn + c = y}}} if {{{y=T[n]}}} where {{{n}}} starts with {{{1}}}, we have

plug in {{{T[1]=4}}}, means first term, {{{n=1}}}
 
{{{a(1)^2 + b + c = 4}}}

{{{a + b + c = 4}}}............. [1]

{{{T[2]=9}}}

{{{a(2)^2 + 2b + c = 9}}}

{{{4a + 2b + c = 9}}}.......... [2]

{{{T[4]=37}}}

{{{a(4)^2 + 4b + c = 37}}}

{{{16a + 4b + c = 37}}}............ [3]

if you subtract [1] from [2], you get

{{{4a + 2b + c-(a + b + c) = 9-4}}}

{{{4a + 2b + c-a -b -c = 5}}}

{{{3a + b  = 5}}}.....solve for {{{b}}}

{{{b  = 5-3a}}}

and [3] - [2] 

{{{16a + 4b + c-(4a + 2b + c) = 37-9}}}

{{{16a + 4b + c-4a -2b - c = 28}}}

{{{12a +2b = 28}}}................plug in {{{b  = 5-3a}}}

{{{12a +2(5-3a) = 28}}}...solve for {{{a}}}

{{{12a +10-6a = 28}}}

{{{6a = 28-10}}}

{{{6a =18}}}

{{{a = 3}}}, ...now find {{{b}}}

{{{b  = 5-3*3}}}

{{{b  = 5-9}}}

so {{{b = -4}}} and .now find {{{c}}}

{{{a + b + c = 4}}}............. [1]

{{{3 -4 + c = 4}}}
{{{-1 + c = 4}}}
{{{c = 4+1}}}
{{{c = 5}}}

then third term will be:

{{{3(3^2) - 4(3) + 2 = x}}}

{{{x = 20 }}}

quadratic pattern:
{{{4}}};{{{9}}};{{{20}}};{{{37}}}

{{{T[n]=3n^2 -4n + 5 }}} where {{{n}}}={{{1}}},{{{2}}},{{{3}}},....