Calculate $\sqrt{4-9x}$ to the power of $2$ and get $4-9x$.
$$4-9x=\left(-1-3x\right)^{2}$$
Use binomial theorem $\left(a-b\right)^{2}=a^{2}-2ab+b^{2}$ to expand $\left(-1-3x\right)^{2}$.
$$4-9x=1+6x+9x^{2}$$
Subtract $1$ from both sides.
$$4-9x-1=6x+9x^{2}$$
Subtract $1$ from $4$ to get $3$.
$$3-9x=6x+9x^{2}$$
Subtract $6x$ from both sides.
$$3-9x-6x=9x^{2}$$
Combine $-9x$ and $-6x$ to get $-15x$.
$$3-15x=9x^{2}$$
Subtract $9x^{2}$ from both sides.
$$3-15x-9x^{2}=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$-9x^{2}-15x+3=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $-9$ for $a$, $-15$ for $b$, and $3$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
