Question 617244
I believe you had a typo in your question. Maybe you had to evaluate ab + bc + ca, and were given as information that
a + b + c = 12 and a*a = b*b +c*c =50
(I'll write that last equation as {{{a*a=b*b+c*c=50}}})
 
If that was the problem, we could start with the product {{{(a+b+c)(a+b+c)}}}
and apply the distributive property twice to get 
{{{(a+b+c)(a+b+c)=a(a+b+c)+b(a+b+c)=c(a+b+c)=a*a+a*b+a*c+b*a+b*b+b*c+c*a+c*b+c*c}}}
To simplify the writing, I would like to start omitting some of those asterisks and middle dot symbols for multiplication,
to be able to write {{{ab}}} instead of a*b or {{{a*b}}}, {{{bc}}} instead of {{{b*c}}}, and the like.
I would also apply the commutative property that says that
{{{ab=a*b= b*a=ba}}}, and similarly {{{ac=ca}}}, and {{{bc=cb}}},
so that I can write all those products as the ab, bc, or ca in the formula that needs to be evaluated
{{{a*a+a*b+a*c+b*a+b*b+b*c+c*a+c*b+c*c=a*a+ab+ca+ab+b*b+bc+ca+bc+c*c}}}
Next, I would rearrange the terms
{{{a*a+ab+ca+ab+b*b+bc+ca+bc+c*c=a*a+b*b+c*c+ab+ab+bc+bc+ca+ca}}}
I would also group them to get together similar terms.
(I also want to group together the expressions that the problem gives values for).
{{{a*a+b*b+c*c+ab+ab+bc+bc+ca+ca=a*a+(b*b+c*c)+((ab+bc+ca)+(ab+bc+ca))}}}
Now we have that
{{{(a+b+c)(a+b+c)=a*a+(b*b+c*c)+((ab+bc+ca)+(ab+bc+ca))}}}
Next, we replace the known values for {{{a+b+c}}} , {{{a*a}}} and {{{b*b+c*c}}}
{{{12*12=50+50+((ab+bc+ca)+(ab+bc+ca))}}} --> {{{144=100+((ab+bc+ca)+(ab+bc+ca))}}} --> {{{144=100+2(ab+bc+ca)}}}
At this point, I first subtract 100 from the expressions on both sides of the equal sign:
{{{144=100+2(ab+bc+ca)}}} --> {{{144-100=100+2(ab+bc+ca)-100}}} --> {{{44=2(ab+bc+ca)}}}
Then, I divide the expressions on both sides of the equal sign by 2:
{{{44=2(ab+bc+ca)}}} --> {{{44/2=2(ab+bc+ca)/2}}} --> {{{22=ab+bc+ca}}}
{{{highlight(ab+bc+ca=22)}}}