How do you find the zeros, real and imaginary, of $y=8x^2-4x-11$ using the quadratic formula?

Question

1680 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-19T12:24:33

$x = (1+-sqrt(23))/4$

For $ax^2+bx+c=0$ :

$x=(-b+-sqrt(b^2-4ac))/(2a)$

Therefore $a=8, b=-4, c=-11$ and:

$x = (4+-sqrt((-4)^2-4*8*-11))/(2*8)$
$x = (4+-sqrt(16+352))/16$
$x = (4+-sqrt(368))/16$
$x = (4+-4sqrt(23))/16$
$x = (1+-sqrt(23))/4$

How do you find the zeros, real and imaginary, of y=8x^2-4x-11y=8x^2-4x-11 using the quadratic formula?