# How do you find the roots, real and imaginary, of y=-4x^2 +x -3 using the quadratic formula?

May 3, 2018

$x = \frac{1 - i \sqrt{47}}{8} ,$ $\frac{1 + i \sqrt{47}}{8}$

The roots are imaginary.

#### Explanation:

$y = - 4 {x}^{2} + x - 3$ is a quadratic equation in standard form:

$y = a {x}^{2} + x - 3$,

where:

$a = - 4$, $b = 1$, $c = - 3$

To find the roots, substitute $0$ for $y$.

$0 = - 4 {x}^{2} + x - 3$

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

Plug in the known values.

$x = \frac{- 1 \pm \sqrt{{1}^{2} - 4 \cdot - 4 \cdot - 3}}{2 \cdot - 4}$

Simplify.

$x = \frac{- 1 \pm \sqrt{- 47}}{- 8}$

$x = \frac{- 1 \pm \sqrt{47 \cdot - 1}}{- 8}$

Simplify $\sqrt{- 1}$ to $i$.

$x = \frac{- 1 \pm i \sqrt{47}}{- 8}$

$47$ is a prime number so it can't be factored.

$x = \frac{1 \pm i \sqrt{47}}{8}$ $\leftarrow$ Two negatives make a positive.

Roots

$x = \frac{1 - i \sqrt{47}}{8} ,$ $\frac{1 + i \sqrt{47}}{8}$

graph{y=-4x^2+x-3 [-15.02, 17, -15.25, 0.77]}