How do you solve 2x^2+6x+33=0?

Oct 23, 2016

$x = - \frac{3}{2} \pm \frac{\sqrt{57}}{2} i$

Explanation:

$2 {x}^{2} + 6 x + 33 = 0$

This is in the form:

$a {x}^{2} + b x + c = 0$

with $a = 2$, $b = 6$ and $c = 33$.

The discriminant $\Delta$ is given by the formula:

$\Delta = {b}^{2} - 4 a c = {\textcolor{b l u e}{6}}^{2} - 4 \left(\textcolor{b l u e}{2}\right) \left(\textcolor{b l u e}{33}\right) = 36 - 264 = - 228 = - {2}^{2} \cdot 57$

Since $\Delta < 0$ this quadratic has a Complex conjugate pair of roots, which we can find using the quadratic formula:

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

$\textcolor{w h i t e}{x} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

$\textcolor{w h i t e}{x} = \frac{- 6 \pm \sqrt{- {2}^{2} \cdot 57}}{2 \cdot 2}$

$\textcolor{w h i t e}{x} = \frac{- 6 \pm 2 \sqrt{57} i}{4}$

$\textcolor{w h i t e}{x} = - \frac{3}{2} \pm \frac{\sqrt{57}}{2} i$