# How do you find the zeros, real and imaginary, of y= x^2-3x-16 using the quadratic formula?

Dec 19, 2017

${x}_{1} = \frac{3 + \sqrt{73}}{2}$ and ${x}_{2} = \frac{3 - \sqrt{73}}{2}$

#### Explanation:

$\Delta = {\left(- 3\right)}^{2} - 4 \cdot 1 \cdot \left(- 16\right) = 73$

Hence solution of y, ${x}_{1} = \frac{3 + \sqrt{73}}{2}$ and ${x}_{2} = \frac{3 - \sqrt{73}}{2}$

Dec 19, 2017

$x = \frac{3 \pm \sqrt{73}}{2}$

#### Explanation:

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

Your equation $y = {x}^{2} - 3 x - 16$ is in standard quadratic form, or in the form of $a {x}^{2} + b x + c$

So...
$a = 1$
$b = - 3$
$c = - 16$

Now let's plug them in the quadratic formula:
$x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(- 16\right)}}{2 \left(1\right)}$
$x = \frac{3 \pm \sqrt{9 + 64}}{2}$
$x = \frac{3 \pm \sqrt{73}}{2}$