# How do you find the zeros, real and imaginary, of y=19x^2-4x-3 using the quadratic formula?

Mar 27, 2018

$x = \frac{2 + \sqrt{41}}{19}$ and $x = \frac{2 - \sqrt{41}}{19}$.

#### Explanation:

The problem says to use the quadratic formula, so we will use that. The quadratic formula is $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$. $a$,$b$, and $c$ come from $a {x}^{2} + b x + c$, the standard form of a quadratic equation.

The equation is already in standard form, so we can simply substitute and simplify!

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
$x = \frac{- \left(- 4\right) \pm \sqrt{{\left(- 4\right)}^{2} - 4 \left(19\right) \left(- 3\right)}}{2 \left(19\right)}$
$x = \frac{4 \pm \sqrt{16 - \left(- 228\right)}}{38}$
$x = \frac{4 \pm \sqrt{244}}{38}$
244 is not a perfect square, so we could stop here. However, notice that 244 is divisible by 4, which is a perfect square, so we can simplify $\sqrt{244}$!
$x = \frac{4 \pm \sqrt{4 \cdot 41}}{38}$
$x = \frac{4 \pm 2 \sqrt{41}}{38}$
We can pull a 2 out of each term and simplify:
$x = \frac{\cancel{2} \left(2 \pm \sqrt{41}\right)}{\cancel{2} \left(19\right)}$
$x = \frac{2 \pm \sqrt{41}}{19}$
So our solutions are $x = \frac{2 + \sqrt{41}}{19}$ and $x = \frac{2 - \sqrt{41}}{19}$.