How do you solve $x^2-4x-2=0$ by using the Quadratic Formula?

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Answer 1 · 2015-07-25T13:52:57

The solutions are:
$color(blue)(x=2+sqrt6$
$color(blue)(x=2-sqrt6$

The equation: $x^2-4x-2$ is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=1, b=-4, c=-2$

The Discriminant is given by:

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

$= (-4)^2 - (4)* (1)*(-2)$
$= 16+8=24$

As $Delta>0$ there are two solutions,

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

$x = (-(-4)+-sqrt(24))/(2*1) = (4+-sqrt(24))/2$

$(color(blue)(sqrt24 = sqrt(2*2*2*3) = 2sqrt6)$

So,
$x= (4+-2sqrt(6))/2$

Taking $2$ outside the bracket as it is common to both terms of the numerator

$x= (cancel2(2+-sqrt(6)))/cancel2$
the solutions are:
$color(blue)(x=2+sqrt6$
$color(blue)(x=2-sqrt6$

How do you solve x^2-4x-2=0x^2-4x-2=0 by using the Quadratic Formula?