# How do you find the zeros, real and imaginary, of y=2x^2-5x-3 using the quadratic formula?

Nov 30, 2017

There are only $2$ real solutions:

$x = 6 \setminus \quad , \setminus \quad x = - 1$

#### Explanation:

First, we can calculate the discriminant to see how many real/imaginary answers there will be:

${b}^{2} - 4 a c$

$\setminus \rightarrow {\left(- 5\right)}^{2} - 4 \left(1\right) \left(- 3\right)$

$\setminus \rightarrow 25 + 24$

$\setminus \rightarrow 49$

$\setminus \therefore$ the equation will have $2$ real answers. There are no imaginary solutions.

Now we can solve for those answers.

First set the RHS equal to $0$:

$2 {x}^{2} - 5 x - 3 = 0$

Now we can just plug the following values into the quadratic formula:

$a = 2$

$b = - 5$

$c = - 3$

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

$\setminus \rightarrow x = \setminus \frac{5 \setminus \pm \setminus \sqrt{{\left(- 5\right)}^{2} - 4 \left(2\right) \left(- 3\right)}}{2 \left(2\right)}$

$\setminus \rightarrow x = \setminus \frac{5 \setminus \pm \setminus \sqrt{25 + 24}}{2}$

$\setminus \rightarrow x = \setminus \frac{5 \setminus \pm \setminus \sqrt{49}}{2}$

$\setminus \rightarrow x = \setminus \frac{5 \setminus \pm 7}{2}$

$\setminus \rightarrow x = \setminus \frac{5 + 7}{2} \setminus \quad , \setminus \quad x = \setminus \frac{5 - 7}{2} 3$

$\setminus \rightarrow x = 6 \setminus \quad , \setminus \quad x = - 1$