How do you solve #x(x-9)+20=0#?

1 Answer
May 13, 2016

Develop the equation

#x^2-9x+20=0#

Your equation is of the form

#ax^2-bx+c=0#

Find the discriminant (#Delta#)

#Delta = b^2-4ac#

#Delta = (-9)^2-4*20= 1#

#Delta# is greater than zero, this means two real roots exist.

#x'= (-b+sqrt(Delta))/(2a)= (-(-9)+ sqrt(1))/2 =5#
#x''= (-b-sqrt(Delta))/(2a)= (-(-9)- sqrt(1))/2 =4#