Question 67968
You can use the distance formula: {{{d = sqrt((x2-x1)^2 + (y2-y1)^2)}}} which is derived from the Pythagorean theorem.
(x1, y1) = (-3, -2) and (x2, y2) = (1, 4)  Making the appropriate substitutions:
{{{d = sqrt((1-(-3))^2 + (4-(-2))^2)}}}
{{{d = sqrt(4^2 + 6^2)}}}
{{{d = sqrt(16+36)}}}
{{{d = sqrt(52)}}}
{{{d = sqrt(4*13)}}}
{{{d = 2sqrt(13)}}}