How do you solve $x^2 + 2x - 8 = 0$ ?

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Answer 1 · 2016-05-21T11:48:08

The solutions for the equation are:
$color(blue)(x = 2$

$color(blue)(x = -4$

$x^2 + 2x - 8 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:
$a=1, b=2, c=- 8$

The Discriminant is given by:

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

$= (2)^2-(4* 1 * (-8))$

$= 4 + 32 = 36$

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

$x = (-2)+-sqrt(36))/(2*1) = ((-2+-6))/2$

$x = (-2 + 6 ) / 2 = 4/2 = 2$ , $color(blue)(x = 2$

$x = (-2 - 6 ) / 2 = -8/2 = -4$ , $color(blue)(x = -4$

How do you solve x^2 + 2x - 8 = 0x^2 + 2x - 8 = 0?