Your answer is as good as any. It is wise to use a method you know will work.
When I multiply the first and last coefficients (2×21), I get a number (42) with factors (-6 and -7) that add to the middle coefficient (-13). As a result, I know the equation can be solved by factoring.
One method of factoring tells you to write the coefficient of the squared term (2) as the leading coefficient in each binomial factor:
(2x + __)×(2x + __) = 0
then fill in the factors of 42 that you found (-6, -7) to get
(2x -6)×(2x -7) = 0
This factorization will give you correct answers (6/2, 7/2), but does not give the original equation when you multiply it out. To get that, you need to remove a factor of 2 (the original leading coefficient). You can do that here by factoring it out of the (2x -6) factor: 2(x -3). (In some instances, you may have to remove factors from both binomials to accomplish this step.)
Then the factoring of the original equation is
(x -3)(2x -7) = 0
At this point, you invoke the "zero product rule," which tells you a product can only be zero if one of its factors is zero. This gives rise to two equations
x - 3 = 0
2x - 7 = 0
The solutions of these are the solutions of your quadratic.
x = 3 or 7/2 . . . . . . these values of x make the factors zero.
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You can also ask a graphing calculator to graph the quadratic and show you the zeros. I like this method because it is easy and only requires I enter the expression into the calculator.