Question 168019
Let consecutive inetgers:
{{{x}}}= 1st integer
{{{x+1}}}=2nd integer
Condition:
{{{(x)(x+1)-20(x+1)=442}}}, right? ------------------> Eqn 1
Continuing,
{{{x^2+x-20x-20-442=0}}}
{{{x^2-19x-462=0}}}
where----{{{system(a=1,b=-19,c=-462)}}}
By Pyth. Theorem:
{{{x=(-b+-sqrt(b^2-4ac))/(2a)}}}
{{{x=(-(-19)+-sqrt(-19^2-4*1*-462))/(2*1)}}}
{{{x=(19+-sqrt(361+1848))/2}}}
{{{x=(19+-sqrt(2209))/2}}}
{{{x=(19+-47)/2}}}
2 values: {{{x=(19+47)/2=66/2=highlight(33)}}}
{{{x=(19-47)/2=-28/2=-14}}}
Use highlighted one=33, 1st integer; 34 = 2nd inetger (Answer)
Go back Eqn 1 to check:
{{{33(33+1)-20(33+1)=442}}}
{{{33*34-20*34=442}}}
{{{1122-680=442}}}
{{{442=442}}}
Thank you,
Jojo