How do you solve $x^2-5x-3=0$ ?

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Answer 1 · 2016-06-20T17:29:52

The solutions are:
$color(blue)(x = ( 5 + sqrt37)/2$

$color(blue)(x = ( 5 - sqrt37)/2$

$x^2 -5x - 3 = 0$

The equation is of the form $color(blue)(ax^2+bx+c=0$ where:

$a=1, b=-5, c=-3$

The Discriminant is given by:

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

$= (-5)^2-(4* 1 * (-3))$

$= 25 + 12 = 37$

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

$x = (-(-5)+-sqrt(37))/(2*1) = (5+-sqrt(37))/( 2 * 1 )$

$x = ( 5 +- sqrt37)/2$

The solutions are:
$color(blue)(x = ( 5 + sqrt37)/2$

$color(blue)(x = ( 5 - sqrt37)/2$

How do you solve x^2-5x-3=0x^2-5x-3=0?