# Find the value of p for which the quadratic equation (p+1)x2-6(p+1)x+3(p+9)=0 had equal roots , hence find the roots

^{2}-4ac

(6p+6)^{2} -4(p+1)(3p+27) =0

36p^{2}^{}+36 +72p -4(3p^{2} +30p +27) =0

36p^{2} +36 +72p-12p^{2}-120p -108 =0

24p^{2}-48p-72 =0

devide by 24

p^{2}-2p-3=0

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

p(p+1)-3(p+1)=0

(p-3)(p+1)=0

p = 3 or -1