How do you solve #x^2 + 5x - 1 = 0#?

1 Answer
Jun 26, 2015

Answer:

Use the standard formula #x = (-b +- sqrt(b^2 - 4ac))/(2a)# to solve an expression #ax^2 + bx + c = 0#

Explanation:

In this case #x =( -5 +- sqrt(5^2 -4*1*(-1)))/(2*1)#
#x =( -5 +- sqrt(25 +4))/2#
#x = (-5 +- sqrt(29))/2#
#x=(-5+sqrt(29))/2 or (-5-sqrt(29))/2#
Using a calculator gives
#x=-5+5.385# or #x=-5-5.385#
#x=0.385# or #x=-10.385#