# How do you solve y = 2x^2 + 3x + 6?

Oct 27, 2015

Assuming you mean:
What are the roots? (find x if y=0)
It does not cross the x-axis so has a complex number solution.
$c = \frac{1}{4} \left(3 \pm \left(\sqrt{39}\right) i\right)$

#### Explanation:

Using standard form equations of:

$y = a {x}^{2} + b x + c \text{ }$ and " x = (-b+-sqrt(b^2 - 4ac))/(2a)

Where a=2; " " b = 3; " and " c =6 giving:

x = (-3 +- sqrt( (3)^2 - 4(2)(6)))/(2(2)

$x = \frac{- 3 \pm \sqrt{9 - 48}}{4}$

$x = - \frac{3}{4} \pm \frac{\sqrt{\left(- 39\right)}}{4}$

But $- 39 = 39 \times \left(- 1\right)$
and $\sqrt{- 1} = i$

giving: $x = \frac{1}{4} \left(3 \pm \left(\sqrt{39}\right) i\right)$