# What is the discriminant of 2x^2 + 5x + 5 = 0 and what does that mean?

Jul 27, 2015

For this quadratic, $\Delta = - 15$, which means that the equation has no real solutions, but it does have two distinct complex ones.

#### Explanation:

The general form for a quadratic equation is

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

The general form of the discriminant looks like this

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

$2 {x}^{2} + 5 x + 5 = 0$

which means that you have

$\left\{\begin{matrix}a = 2 \\ b = 5 \\ c = 5\end{matrix}\right.$

The discriminant will thus be equal to

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

$\Delta = 25 - 40 = \textcolor{g r e e n}{- 15}$

The two solutions for a general quadratic are

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

When $\Delta < 0$, such as you have here, the equation is said to have no real solutions, since you're extracting the square root from a negative number.

However, it does have two distinct complex solutions that have the general form

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

In your case, these solutions are

${x}_{1 , 2} = \frac{- 5 \pm \sqrt{- 15}}{4} = \left\{\begin{matrix}{x}_{1} = \frac{- 5 + i \sqrt{15}}{4} \\ {x}_{2} = \frac{- 5 - i \sqrt{15}}{4}\end{matrix}\right.$