What is a quadratic equation in standard form with rational coefficients that has a root of 5 + 4i?

Help with solution to 2 variable quadratic equation.
Could someone help with an explanation on how to treat the equation below, please? A solution I have read uses the discriminant to find the range i.e.
${\displaystyle 9-20y^2-16y>\text{or}=0}$. To me this means treating the variable Y as a constant , is that "acceptable" ? does anyone have a reference for these equations ?
${\displaystyle y{x}^2+5y-3x+4=0}$
rewrite as ${\displaystyle y{x}^2-3x+(5y+4)=0}$

Making a Negative Number Possible to Square Root
We are able to solve ${\displaystyle x^2+4=0}$ by square rooting both sides, but if we have ${\displaystyle x^2=-4}$ we can't solve. Firstly, why? Aren't they equal expressions?
Secondly, if we have ${\displaystyle x^2=-4}$, why can't we bring the four to the other side, square root it and then bring it back?

Let ${\displaystyle x^2-(m-3)x+m=0,(m\in R)}$ a quadratic equation. Find the value of m for which at least one root is greater than 2

Quadratic equation
${\displaystyle 3x^3=24}$

I need some help to solving a quadratic equation, but translating it into Dedekind Cuts and using the completing the square way. I first solved the equation in a usual way so i can know where do i have to get, but the Dedekind Cut stuff, I don't know how to use it on it.
Given: ${\displaystyle x^2+6x+3=0}$

Quadratic inequality with Modulus of a Linear Equation : Solution Verification & Clarification Needed
${\displaystyle x^2-|5x-3|-x<2}$
For this problem I took two cases where in one case ${\displaystyle 5x-3\ge 0}$ and in second case ${\displaystyle 5x-3<0}$
1) ${\displaystyle 5x-3\ge 0}$:
2) ${\displaystyle 5x-3<0}$

How to solve a quadratic equation with Binary coefficients?
${\displaystyle 5x^2-50x+125=0}$
It has the roots ${\displaystyle x_1=5}$ and ${\displaystyle x_2=5}$. But now, convert these coefficients into binary:
${\displaystyle 101x^2-110010x+1111101=0}$
Solve this quadratic equation with binary coefficients.