find the value of A if the equation (a+3)x²-(11a+1)x+a=2(a-5) they are equal root

Answer 1

(a+3)x^2 - (11a+1)x + a - 2(a-5) = 0

(a+3)x^2 - (11a+1)x + (10-a) = 0

discrimminant is (11a+1)^2 - 4(a+3)(10-a) =

(11a+1)^2 - 4(a+3)(10-a) =

(11a+1)^2 +4(a+3)(a-10) =

(11a +1)^2 - 4( a^2 - 7a - 30) =

121a^2 + 22a + 1 - 4a^2 + 28a + 120 =

117a^2 + 50a + 121