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

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Answer 1 · 2016-05-16T17:05:10

The solutions are:
$x = - 1/2$

$x = - 1$

$2x^2 + 3x + 1 = 0$
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:

$Delta=b^2-4*a*c$

$= (3)^2-(4 * 2 * 1)$

$= 9 - 8 = 1$

The solutions are found using the formula
$x=(-b+-sqrtDelta)/(2*a)$

$x = ((-3)+-sqrt(1))/(2*2) = (-3 +- ( 1) )/4$

$x = (- 3 + 1 ) / 4 = -2/ 4 = - 1/2$

$x = (- 3 - 1 ) / 4 = -4/4 = - 1$

How do you solve 2x^2+3x+1=02x^2+3x+1=0?