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

Question

56685 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-06-12T08:31:23

the solutions are
$x = 3$ $color(green)(x= 3$
$x = - 1$ $color(green)( x=-1$

$x^{2} - 2 x - 3 = 0$ $x^2 - 2x - 3 = 0$

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

The Discriminant is given by:
$Delta=b^2-(4*a*c)$

$= (-2)^2-(4*1*(-3))$
$= 4 + 12= 16$

As $Delta>0$ there are two solutions,

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

As $Delta = 16$ , $x = (-(-2)+-sqrt(16))/(2*1) = (2+-4)/2$

the solutions are
$color(green)(x= 6/2 = 3$
$color(green)( x=-2/2=-1$

How do you solve x^2 - 2x - 3 = 0x2−2x−3=0x^2 - 2x - 3 = 0 using the quadratic formula?