Question

How do you solve using the quadratic formula for `x^2 + x + 5 = 0`?

Answer 1

The answer is `(-1+-isqrt(19))/2`.

Explanation:

The quadratic formula is `x=(-b+-sqrt(b^2-4ac))/(2a` for the equation `ax^2+bx+c`.

In this case, `a=1`, `b=1`, and `c=5`.

You can therefore substitute in those values to get:

`(-1+-sqrt(1^2-4(1)(5)))/(2(1)`.

Simplify to get `(-1+-sqrt(-19))/2`.

Because `sqrt(-19)` is not a real number, we have to stick to imaginary solutions. (If this problem asks for real number solutions, there are none.)

The imaginary number `i` equals `sqrt(-1)`, therefore we can substitute it in:

`(-1+-sqrt(-1*19))/2 rarr (-1+-sqrt(-1)*sqrt(19))/2 rarr (-1+-isqrt(19))/2`, the final answer.

Hope this helps!

Answer 2

See application of the quadratic formula below in obtaining the result:
`color(white)("XXX")x=-1/2+-sqrt(19)i`

Explanation:

`x^2+x+5=0` is equivalent to `color(red)1x^2+color(blue)1x+color(magenta)5=0`

Applying the general quadratic formula `x=(-color(blue)b+-sqrt(color(blue)b^2-4color(red)acolor(magenta)c))/(2color(red)a`
for `color(red)ax^2+color(blue)bx+color(magenta)c=0`

to this specific case, we have
`color(white)("XXX")x=(-color(blue)1+-sqrt(color(blue)1^2-4 * color(red)1 * color(magenta)5))/(2 *color(red)1)`

`color(white)("XXXXX")=(-1+-sqrt(-19))/2`

There are no Real solutions, but as Complex values:
`color(white)("XXX")x=-1/2+sqrt(19)icolor(white)("XXX")"or"color(white)("XXX")x=-1/2-sqrt(19)i`