Quadratic equation with real number coefficients

[raw iron]

I don't know what to do with the solution of 6-7i

[umbral shard]

one of the linear factors of your quadratic function over C must be (x - (6 - 7i))

[raw iron]

oh ok

thank you

[smoky island]

so basically, if a solution is imaginary

you can know a second solution

which is the conjugate zero

if you don't know what a conjugate zero is, it's basically the same number, except the term with i is multiplied by -1

(sorry im not sure if i explained this well) but basically if a solution is a+bi, there is another solution that is a-bi.

a very simple proof for this is that in the quadratic formula, the only place an imaginary number can form is if the part in the root (b^2-4ac) is negative, and the two solutions to a quadratic equation is -b+√b^2-4ac, and -b-√b^2-4ac

this means that there is always a conjugate zero

so basically, knowing the two zeros

the factored version of the equation is (x-(6-7i))(x-(6+7i))

and basically i cancels out and the equation becomes x^2-12x+85=0