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

Question

4819 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-12T08:11:51

#x=0.171# or #x=-1.171#

#5x^2+5x-1=0#

Standard form of a quadratic equation:

#ax^2+bx+c=0#

#x=(-b+sqrt(b^2-4ac))/(2a)#

#:.x=((-5)+-sqrt(5^2-(4*5*-1)))/(2*5)#

#:.x=(-5+-sqrt(25+20))/10#

#:.x=(-5+sqrt45)/10# or #x=(-5-sqrt45)/10#

#:.x=(-5+6.708203933)/10# or #x=(-5-6.708203933)/10#

#:.x=1.708203933/10# or #x=-11.708203933/10#

#:.x=0.1708203933# or #x=-1.1708203933#

#:.x=0.171# or #x=-1.171#