How do you solve $x^2 + 3x + 6 = 0$ by quadratic formula?

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Answer 1 · 2015-12-30T12:04:55

Substitute the coefficients $a=1$ , $b=3$ and $c=6$ into the quadratic formula to find:

$x=(-3+-i sqrt(15))/2$

$x^2+3x+6$ is in the form $ax^2+bx+c$ with $a=1$ , $b=3$ and $c=6$ .

First note that the discriminant $Delta$ is negative.

$Delta = b^2-4ac = 3^2-(4xx1xx6) = 9 - 24 = -15$

So our quadratic equation has two Complex roots.

The roots are given by the quadratic formula:

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

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

$=(-3+-sqrt(-15))/2$

$=(-3+-i sqrt(15))/2$

Answer 2 · 2015-12-30T12:07:05

The solutions are:
$x= color(blue)((-3+sqrt(-15))/2$

$x= color(blue)((-3-sqrt(-15))/2$

$x^2 +3x +6 =0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=1, b=3, c=6$

The Discriminant is given by:
$Delta=b^2-4*a*c$

$= (3)^2-(4*(1)*6)$

$= 9-24 = -15$

As $Delta = -15$ , this equation has NO REAL SOLUTIONS

The solutions are found using the formula:
$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-3)+-sqrt(-15))/(2*1) = (-3+-sqrt(-15))/2$

The solutions are:
$x= color(blue)((-3+sqrt(-15))/2$

$x= color(blue)((-3-sqrt(-15))/2$

How do you solve x^2 + 3x + 6 = 0x^2 + 3x + 6 = 0 by quadratic formula?