Find the set of values of x for which 3x^2+8x-3<0.

This is a question involving inequalities. An inequality is a statement about real numbers involving one of the symbols >, ≥, ≤, <. In our question we have to find when function f(x)=3x^2+8x-3 is less than 0. The best way to see when f(x) is smaller than 0 is to draw a graph of f(x). We know that f(x) is a quadratic function because it has a form ax^2+bx+c. First, we will find roots of f(x). This can be done by substituting the values into quadratic formula (-b±sqrt(b^2-4ac))/2a. So we'll get (-8±sqrt(8^2-4·3·(-3))), which will give us roots x=1/3 and x=-3. A quadratic equation where the value a is positive has a shape U. So the graph will start from the second quadrant. It will go down through x-axis where x=-3 and then go up through the point x=1/3 and continue going up in the first quadrant. The part of a graph where y values are smaller then 0 is the set of values of x that satisfies the inequality. These values are when x is between -3 and 1/3. However, we have to be careful not to include values x=-3 and x=1/3 because there is a symbol < not ≤. So the final answer is (-3,1/3) (notice we are using round brackets to describe an open interval).