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

Dec 30, 2015

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

$x = \frac{- 3 \pm i \sqrt{15}}{2}$

#### Explanation:

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

First note that the discriminant $\Delta$ is negative.

$\Delta = {b}^{2} - 4 a c = {3}^{2} - \left(4 \times 1 \times 6\right) = 9 - 24 = - 15$

So our quadratic equation has two Complex roots.

The roots are given by the quadratic formula:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

$= \frac{- b \pm \sqrt{\Delta}}{2 a}$

$= \frac{- 3 \pm \sqrt{- 15}}{2}$

$= \frac{- 3 \pm i \sqrt{15}}{2}$

Dec 30, 2015

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

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

#### Explanation:

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

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:
$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(3\right)}^{2} - \left(4 \cdot \left(1\right) \cdot 6\right)$

$= 9 - 24 = - 15$

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

The solutions are found using the formula:
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 3\right) \pm \sqrt{- 15}}{2 \cdot 1} = \frac{- 3 \pm \sqrt{- 15}}{2}$

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

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