# How do you find the discriminant for x^2-4/5x=3 and determine the number and type of solutions?

Nov 21, 2017

See a solution process below:

#### Explanation:

First, rewrite the equation in standard form as:

${x}^{2} - \frac{4}{5} x - 3 = 0$

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:

$\textcolor{red}{1}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{- \frac{4}{5}}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{3}$ for $\textcolor{g r e e n}{c}$

${\textcolor{b l u e}{- 3}}^{2} - \left(4 \cdot \textcolor{red}{1} \cdot \textcolor{g r e e n}{- \frac{4}{5}}\right)$

$9 - \left(- \frac{16}{5}\right)$

$9 + \frac{16}{5}$

$9 + \frac{15}{5} + \frac{1}{5}$

$9 + 3 + \frac{1}{5}$

$12 + \frac{1}{5}$

$12 \frac{1}{5}$ or $\frac{61}{5}$

Because the discriminate is positive there will two (2) real solutions for this equation.