If one root of the equation $x^2+px+12=0$  is 4, while the equation $x^2+px+q=0$ has equal roots, then the value of ‘q’ is

$16+4p+12=0$

$4p=-28$

$p=-7$

Given $x^2+px+q=0$  has equal roots

$x^2-7x+q=0$  has equal roots

$b^2-4ac=0$

$49-4q=0$

$q=\frac{49}{4}$