# How to use the discriminant to find out how many real number roots an equation has for 9n^2 - 3n - 8= -10?

Jul 5, 2015

There is no real number root to $9 {n}^{2} - 3 n - 8 = - 10$

#### Explanation:

The first step is to change the equation to the form:

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

To do so, you must do:

$9 {n}^{2} - 3 n - 8 + 10 = - \cancel{10} + \cancel{10}$
$\rightarrow 9 {n}^{2} - 3 n + 2 = 0$

Then, you must calculate the discriminant:

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

$a = 9$
$b = - 3$
$c = 2$

Therefore:

$\Delta = {\left(- 3\right)}^{2} - 4 \cdot 9 \cdot 2 = 9 - 72 = - 63$

Depending on the result, you can conclude how many real solutions exist:

if $\Delta > 0$, there are two real solutions:
$\rightarrow {n}_{+} = \frac{- b + \sqrt{\Delta}}{2 a}$ and ${n}_{-} = \frac{- b - \sqrt{\Delta}}{2 a}$

if $\Delta = 0$, there is one real solution:
$\rightarrow {n}_{0} = \frac{- b}{2 a}$

if $\Delta < 0$, there is no real solution.

In your case, $\Delta = - 63 < 0$, therefore there is no real number root to $9 {n}^{2} - 3 n - 8 = - 10$