How do you solve $4 x^2 – 3x + 3 = 0$ using the quadratic formula?

Question

1946 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-01T17:07:53

The solutions for the equation are :
$=(3+sqrt(-39))/8$
$=(3-sqrt(-39))/8$

$4x^2 -3x + 3 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=4, b=-3, c=3$

The Discriminant is given by:

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

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

$= 9 - 48=- 39$

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

$x = (-(-3)+-sqrt(-39))/(2*4) = (3+-sqrt(-39))/8$

The solutions are:
$=(3+sqrt(-39))/8$
$=(3-sqrt(-39))/8$

How do you solve 4 x^2 – 3x + 3 = 04 x^2 – 3x + 3 = 0 using the quadratic formula?