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How many solutions per each equation?

How many solutions per each equation?-example-1

1 Answer

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assuming y=ax²+bx+c, calculate D = b²-4ac, D<1: no solutions; D=0: 1 solution; D>0: 2 solutions.

1-4*-3*12 = 145, so 2 solutions
36-4*2*5 = -4, no solutions
49-4*-11 = 5, so 2 solutions
64-4*-1*-16 = 0, so 1 solution
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