# How do you solve 2x^2+3x+1=0?

May 16, 2016

The solutions are:
$x = - \frac{1}{2}$

$x = - 1$

#### Explanation:

$2 {x}^{2} + 3 x + 1 = 0$
4
The equation is of the form color(blue)(ax^2+bx+c=0 where:

$a = 2 , b = 3 , c = 1$

The Discriminant is given by:

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

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

$= 9 - 8 = 1$

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

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

$x = \frac{- 3 + 1}{4} = - \frac{2}{4} = - \frac{1}{2}$

$x = \frac{- 3 - 1}{4} = - \frac{4}{4} = - 1$