Question 33680
Start with the formula for the area of a triangle:
{{{A = (1/2)bh}}} Where: A = 60, h = b-2
{{{60 = (1/2)b(b-2)}}} Simplify and solve for b.
{{{60 = (1/2)(b^2-2b)}}} Multiply both sides by 2.
{{{120 = b^2-2b}}} Subtract 120 from both sides of the equation.
{{{b^2-2b-120 = 0}}} Solve this quadratic for b by factoring.
{{{(b+10)(b-12) = 0}}} Apply the zero products principle.
{{{b+10 = 0}}} and/or {{{b-12 = 0}}}
If {{{b+10 = 0}}} then {{{b = -10}}} Discard this solution as b can only be a positive value.
If {{{b-12 = 0}}} then {{{b = 12}}} This solution is acceptable.

The length of the base is 12 mm.
The height would be 10 mm 

Check:
{{{A = (1/2)bh}}}
{{{A = (1/2)(12)(10)}}}
{{{A = (1/2)(120)}}}
{{{A = 60}}} sq.mm.