Question 27769
Let the two numbers be a and b.
You can write two equations from the problem description.

{{{a + b = 12}}} Solve this one for a and substitute into the second equation. {{{a = 12-b}}}
{{{3ab = 105}}}

{{{3(12-b)b = 105}}} Simplify and solve for b.
{{{36b - 3b^2 = 105}}} Subtract 105 from both sides of the equation.
{{{36b - 3b^2 - 105 = 0}}} Multiply through by -1.
{{{3b^2 - 36b + 105 = 0}}} Factor out a 3.
{{{3(b^2 - 12b + 35) = 0}}} Apply the zero products principle.
{{{b^2 - 12b + 35 = 0}}} Factor this quadratic equation.
{{{(b - 7)(b - 5) = 0}}}  Apply the zero products principle again.
{{{b - 7 = 0}}} and/or {{{b - 5 = 0}}}
If {{{b - 7 = 0}}} then {{{b = 7}}}
If {{{b - 5 = 0}}} then {{{b = 5}}}

The two numbers are: 5 and 7