How do you solve #p^ { 2} - 29p + 80= 0#?

Question

1015 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-08T22:54:05

#p_"1"=3.08#
#p_"2"=25.9125#

its a second degree polynomial and can be solved like:

let a be coefficient of the #p^2# which is 1
let b be coefficient of the #p^1# which is -29
let c be coefficient of the #p^0# which is +80

now let's find #Delta# by

#Delta=b^2-4ac#

#Delta=(-29)^2-4*1*80 = 841-320=521#

#Delta=521#

since #Delta>0# there are 2 real roots
now lets find them (#p_"1",p_"2"#)

#p_"1"=(-b-sqrt(Delta))/(2a)# ,

#p_"2"=(-b+sqrt(Delta))/(2a)#

#p_"1"=(-(-29)-sqrt(521))/(2) =(29-22.825)/(2) = 3.08 #

#p_"2"=(-(-29)+sqrt(521))/(2) =(29+22.825)/(2) = 25.9125 #