How do you find the zeros, real and imaginary, of $y=x^2-8x-3$ using the quadratic formula?

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Answer 1 · 2016-06-09T15:59:09

This function hs $2$ real zeros:

$x_1=4-sqrt(19)$

$x_2=4+sqrt(19)$

To calculate the zeros of a quadratic function first you calculate the determinant:

$y=x^2-8x-3$

$Delta = b^2-4ac=8^2-4xx1xx(-3)=64+12=76$

$sqrt(Delta)=sqrt(76)=sqrt(4xx19)=2sqrt(19)$

$Delta >=0$ , so the function has $2$ real zeros. (If $Delta<0$ then the zeros would be complex numbers)

$x_1=(-b-sqrt(Delta))/(2a)$

$x_1=(8-2sqrt(19))/(2)=4-sqrt(19)$

$x_2=(-b+sqrt(Delta))/(2a)$

$x_1=(8+2sqrt(19))/(2)=4+sqrt(19)$

How do you find the zeros, real and imaginary, of y=x^2-8x-3y=x^2-8x-3 using the quadratic formula?