How do you solve $2x^2-3x-1=0$ using the quadratic formula?

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Answer 1 · 2015-08-10T07:23:10

The solutions for the equation are:

$color(blue)(x=(3+sqrt17)/4)$

$color(blue)(x=(3-sqrt17)/4)$

$2x^2−3x−1=0$

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:
$Delta=b^2-4*a*c$
$= (-3)^2-(4*(2)*-1)$
$= 9+8 = 17$

The solutions are found using the formula

$x=(-b+-sqrtDelta)/(2*a)$

$x = (-(-3)+-sqrt(17))/(2*2) = (3+-sqrt(17))/4$

$color(blue)(x=(3+sqrt17)/4)$

$color(blue)(x=(3-sqrt17)/4)$

How do you solve 2x^2-3x-1=02x^2-3x-1=0 using the quadratic formula?