Solve y=ax^2+4x

$$y = a x ^ { 2 } + 4 x$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{y-4x}{x^{2}}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&y=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$ax^{2}+4x=y$$

Subtract 4x from both sides.

$$ax^{2}=y-4x$$

The equation is in standard form.

$$x^{2}a=y-4x$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{y-4x}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$a=\frac{y-4x}{x^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{y-4x}{x^{2}}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&y=0\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$ax^{2}+4x=y$$

Subtract 4x from both sides.

$$ax^{2}=y-4x$$

The equation is in standard form.

$$x^{2}a=y-4x$$

Divide both sides by x^{2}.

$$\frac{x^{2}a}{x^{2}}=\frac{y-4x}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$a=\frac{y-4x}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{ay+4}+2}{a}\text{; }x=\frac{\sqrt{ay+4}-2}{a}\text{, }&a\neq 0\\x=\frac{y}{4}\text{, }&a=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{ay+4}+2}{a}\text{; }x=\frac{\sqrt{ay+4}-2}{a}\text{, }&\left(y\leq -\frac{4}{a}\text{ and }a<0\right)\text{ or }\left(y\geq -\frac{4}{a}\text{ and }a>0\right)\text{ or }\left(a\neq 0\text{ and }y=-\frac{4}{a}\right)\\x=\frac{y}{4}\text{, }&a=0\end{matrix}\right.$$