How do you solve 2x^2 + x + 7 = 0?

Mar 9, 2016

The solutions are:
$x = \frac{- 1 + \sqrt{- 55}}{4}$

$x = \frac{- 1 - \sqrt{- 55}}{4}$

Explanation:

$2 {x}^{2} + x + 7 = 0$

The equation is of the form color(blue)(ax^2+bx+c=0 where:
$a = 2 , b = 1 , c = 7$

The Discriminant is given by:

$\Delta = {b}^{2} - 4 \cdot a \cdot c$

$= {\left(1\right)}^{2} - \left(4 \cdot 2 \cdot 7\right)$

$= 1 - 56 = - 55$

The solutions are found using the formula
$x = \frac{- b \pm \sqrt{\Delta}}{2 \cdot a}$

$x = \frac{\left(- 1\right) \pm \sqrt{- 55}}{2 \cdot 2} = \frac{- 1 \pm \sqrt{- 55}}{4}$

The solutions are:
$x = \frac{- 1 + \sqrt{- 55}}{4}$

$x = \frac{- 1 - \sqrt{- 55}}{4}$