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

Nov 27, 2015

#### Answer:

$x = + 1 \frac{1}{2} \textcolor{w h i t e}{\ldots \ldots \ldots} x = - 1$

$\left\{\exists x : \forall x \in \mathbb{R}\right\} \textcolor{w h i t e}{\ldots .}$ Calculating y is just a matter of substituting for $x$. However it is obvious that $y = 0$

#### Explanation:

Given:$\textcolor{w h i t e}{. .} y = - 2 {x}^{2} + x + 3$

Standard form equation: $\textcolor{w h i t e}{. .} y = a {x}^{2} + b x + c$

$\textcolor{w h i t e}{\ldots . .} x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

In this case:

$a = \left(- 2\right)$
$b = 1$
$c = 3$

So$\textcolor{w h i t e}{\ldots .} x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \left(- 2\right) \left(3\right)}}{2 \left(- 2\right)}$

$\textcolor{w h i t e}{\ldots .} x = \frac{- 1 \pm \sqrt{1 + 24}}{- 4}$

$x = \frac{- 1 \pm 5}{- 4}$

$x = + 1 \frac{1}{2} \textcolor{w h i t e}{\ldots \ldots \ldots} x = - 1$