# How do you find the value of the discriminant and determine the nature of the roots -4r^2-4r=6?

Jul 24, 2017

See a solution process below:

#### Explanation:

First, put this equation in standard form:P

$- 4 {r}^{2} - 4 r - \textcolor{red}{6} = 6 - \textcolor{red}{6}$

$- 4 {r}^{2} - 4 r - 6 = 0$

The quadratic formula states:

For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:

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

The discriminate is the portion of the quadratic equation within the radical: ${\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}$

If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions

To find the discriminant for this problem substitute:

$\textcolor{red}{- 4}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- 4}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{- 6}$ for $\textcolor{g r e e n}{c}$

Giving:

${\textcolor{b l u e}{- 4}}^{2} - \left(4 \cdot \textcolor{red}{- 4} \cdot \textcolor{g r e e n}{- 6}\right)$

$16 - 96$

$- 80$

Because the Discriminate is negative you get a complex solution.