Find two additional roots of p(x)=0. 13+sqrt 3 and 7i

So, x = 13, x = √3 and x =7i.

now, recall that for an EVEN radical, there are two possible roots, namely is say √3 is say hmmm some value "a", that means that a*a = √3, however, -a*-a is also √3, therefore, ±√3 are two valid values, and therefore -√3 is another one.

now keep in mind that, complex solutions or roots, never come all by their lonesome, their sister is always with them, the conjugate, so, for 7i or namely 0 + 7i, her sister is always around, 0 - 7i, which is the other root.

n=48

Step-by-step explanation:

add 48 to both sides to get rid of it43+5=48 n=48