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

Question

13456 views around the world

You can reuse this answer
Creative Commons License

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#