How do you find the zeros, real and imaginary, of $y=8x^2-4x-73$ using the quadratic formula?

Question

1337 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-06T11:53:56

$x approx 3.281 or -2.781$

$y=8x^2-4x-73$

Let's first check the discriminant:
$Delta = (-4)^2 - 4xx8xx(-73)$

$= 16+2336 = 2352$

Since $Delta >0$ $y$ will have 2 real roots.

Apply the quadratic fomrula:

$x= (4+-sqrt(2352))/(2xx8)$

$approx (4+-48.4874)/16$

$approx 3.281 or -2.781$

How do you find the zeros, real and imaginary, of y=8x^2-4x-73y=8x^2-4x-73 using the quadratic formula?