二次方程功課問題

if x=3+4j is a root of x^2+ax+b=0,find the real numbers a and b
唔係用呢條equation....x= -b+ - root b^2 -4ac/2a咩....
我見到其他地方有answer不過唔明點計.......計黎計去都唔得....請問有無人話我知
x^2 + ax + b = 0

For x = 3 + 4j is a root, then its conjugate x = 3 - 4j must be another root.

Sum of roots, -a = (3 + 4j) + (3 - 4j) = 6

So, a = -6

Product of roots, b = (3 + 4j)(3 - 4j) = (3)^2 - (4j)^2 = 9 - (-16) = 25

So, a = -6, b = 25