I'm confused.. I need help ASAP

Let one solution of a quadratic function be x=5i. Find the quadratic function. What is the "C" value of the quadratic.

\(0=k(ax^2+bx+c)\\ \)

\(x = {-b + \sqrt{b^2-4ac} \over 2a}=5i\\ so\;\;b=0\\ { + \sqrt{-4ac} \over 2a}=5i\\ { +2 \sqrt{-ac} \over 2a}=5i\\ \sqrt{-ac}=5ai\\ -ac=-25a^2\\ c=25a\\ \)

I can effectively ignore the k because it will not change the relationship between a and c

There is probably a much quicker way to to this. :/

If one root is 5i,then since complex roots occur in conjugate pairs, we have

quadratic  = (x-5i)(x+5i)  which is identical to ax^2 +bx +c

expanding left hand side and comparing co-efficients

x^2 +x(5i -5i) -25i^2    = ax^2 +bx +c

x^2 +x(0) +25               = ax^2 +bx =c

so a = 1    b=0  and c= 25

typo error,apologies,    b does NOT equal 0,   -- co-efficient of b is zero.

err,sorry again...bad hair day   , b is indeed zero. More coffee needed.