How do you find the zeros, real and imaginary, of y=x^2-3x+298 using the quadratic formula?

Mar 9, 2016

$x = \frac{3}{2} + \frac{13 \sqrt{7}}{2} i$

$x = \frac{3}{2} - \frac{13 \sqrt{7}}{2} i$

Explanation:

Standard form:$\text{ } y = a {x}^{2} + b x + c$

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

$a = 1$
$b = - 3$
$c = 298$

$\implies x = \frac{- \left(- 3\right) \pm \sqrt{{\left(- 3\right)}^{2} - 4 \left(1\right) \left(298\right)}}{2 \left(1\right)}$

$\implies x = \frac{3 \pm \sqrt{- 1183}}{2}$

'~~~~~~~~~~~~~~~~~~~~~
$169 \times 7 = 1183$
$169 = {13}^{2}$
'~~~~~~~~~~~~~~~~~~~
$\implies x = \frac{3 \pm \sqrt{- 1 \times 7 \times {13}^{2}}}{2}$

$\implies x = \frac{3 \pm 13 \sqrt{- 1} \sqrt{7}}{2}$

$x = \frac{3}{2} + \frac{13 \sqrt{7}}{2} i$

$x = \frac{3}{2} - \frac{13 \sqrt{7}}{2} i$