# What is the discriminant of 3x^2 + 6x + 5 and what does that mean?

Jul 27, 2015

For this quadratic, $\Delta = - 24$, which means that the equation has no real solution, but that it does have two distinct complex ones.

#### Explanation:

For a quadratic equation written in general form

$a {x}^{2} + b x + c = 0$,

the discriminant is defined as

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

$3 {x}^{2} + 6 x + 5 = 0$,

which means that you have

$\left\{\begin{matrix}a = 3 \\ b = 6 \\ c = 5\end{matrix}\right.$

The discriminant will thus be equal to

$\Delta = {6}^{2} - 4 \cdot 3 \cdot 5$

$\Delta = 36 - 60 = \textcolor{g r e e n}{- 24}$

When $\Delta < 0$, the equation has no real solutions. It does have two distinct complex solutions derived from the general form

${x}_{1 , 2} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

which in this case becomes

${x}_{1 , 2} = \frac{- b \pm i \sqrt{- \Delta}}{2 a}$, when $\Delta < 0$.

In your case, these two solutions are

${x}_{1 , 2} = \frac{- 6 \pm \sqrt{- 24}}{2 \cdot 3}$

${x}_{1 , 2} = \frac{- 6 \pm i \sqrt{24}}{6} = \frac{- 6 \pm 2 i \sqrt{6}}{6} = \left\{\begin{matrix}{x}_{1} = \frac{- 3 - i \sqrt{6}}{3} \\ {x}_{2} = \frac{- 3 + i \sqrt{6}}{3}\end{matrix}\right.$