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

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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 ( $p_"1",p_"2"$ )

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

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

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

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

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