How do you find the zeros, real and imaginary, of $y=3x^2-17x-9$ using the quadratic formula?

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Answer 1 · 2015-12-25T13:43:23

$x_1 = (17 - sqrt397)/6$ and $x_2 = (17 + sqrt397)/6$

You first need to calculate $b^2 - 4ac = Delta$ . Here, $Delta = 289 + 4*3*9 = 289 + 108 = 397 > 0$ so it has 2 real roots.

The quadratic formula tells us that the roots are given by $(-b +- sqrtDelta)/(2a)$ .

$x_1 = (17 - sqrt397)/6$ and $x_2 = (17 + sqrt397)/6$

How do you find the zeros, real and imaginary, of y=3x^2-17x-9y=3x^2-17x-9 using the quadratic formula?