Question 805355
{{{(p+3)x^2-12x+2p=0}}}


In any quadratic equation if the roots are equal then {{{b^2-4ac) = 0}}} in the equation {{{ax^2+bx+c}}}

compare the two equations

a=(p+3), b=-12, c=2p

{{{b^2-4ac=0}}}

(-12)^2-4(p+3)(2p)=0

144-8p^2-24p=0

Or

8p^2+24p-144=0

/8
{{{p^2+3p-18=0}}}

{{{p^2+6p-3p-18=0}}}


p(p+6)-3(p+6)=0

(p+6)(p-3)=0

p=-6 OR 3