How do you find the zeros, real and imaginary, of $y= x^2-3x-16$ using the quadratic formula?

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Answer 1 · 2017-12-19T23:15:53

$x_1=(3+sqrt73)/2$ and $x_2=(3-sqrt73)/2$

$Delta=(-3)^2-4*1*(-16)=73$

Hence solution of y, $x_1=(3+sqrt73)/2$ and $x_2=(3-sqrt73)/2$

Answer 2 · 2017-12-19T23:30:18

$x = (3 +- sqrt(73))/2$

The quadratic formula is $x = (-b+- sqrt(b^2 - 4ac))/(2a)$

Your equation $y = x^2 - 3x - 16$ is in standard quadratic form, or in the form of $ax^2 + bx + c$

So...
$a = 1$
$b = -3$
$c = -16$

Now let's plug them in the quadratic formula:
$x = (-(-3) +- sqrt((-3)^2 - 4(1)(-16)))/(2(1))$
$x = (3 +- sqrt(9 + 64))/2$
$x = (3 +- sqrt(73))/2$

How do you find the zeros, real and imaginary, of y= x^2-3x-16y= x^2-3x-16 using the quadratic formula?