Question

3x²-12x+(n-5)=0; find the value of n.

Answer 1

3x²-12x+(n-5)=0
Here a=3,b=-12,c=n-5
Given= Equation has equal roots
Then, D=0
b²-4ac=0
(-12)²-4(3)(n-5)=0
144-12n+60=0
-12n=-204
n= 17
Hope this helps

Answer 2

$3 {x}^{2} - 12x + (n - 5) = 0 \\ d = \sqrt{ {( - 12)}^{2} - 4(3)(n - 5) } \\ for \: equal \: roots \: d = 0 \\ 144 = 4(3)(n - 5) \\ 12 = n - 5 \\ n = 17......(1)$
since for d>=0 n could get any value after n=17



mark as brainliest if helped