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For the equation x2 + 3x + j = 0, find all the values of j such that the equation has two real number solutions. Show your work.
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Hint: D = b^2 - 4ac is the discriminant and has 2 real solutions when D > 0 D > 0 b^2 - 4ac > 0 In this case, a = 1, b = 3 and c = j, so... 3^2 - 4*1*j > 0 9 - 4j > 0 Keep going to solve for j
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