Solve y=Ax^2+40

$$y = A x ^ { 2 } + 40$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{y-40}{x^{2}}\text{, }&x\neq 0\\A\in \mathrm{C}\text{, }&y=40\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$Ax^{2}+40=y$$

Subtract 40 from both sides.

$$Ax^{2}=y-40$$

The equation is in standard form.

$$x^{2}A=y-40$$

Divide both sides by x^{2}.

$$\frac{x^{2}A}{x^{2}}=\frac{y-40}{x^{2}}$$

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

$$A=\frac{y-40}{x^{2}}$$

Solve for A

$$\left\{\begin{matrix}A=\frac{y-40}{x^{2}}\text{, }&x\neq 0\\A\in \mathrm{R}\text{, }&y=40\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$Ax^{2}+40=y$$

Subtract 40 from both sides.

$$Ax^{2}=y-40$$

The equation is in standard form.

$$x^{2}A=y-40$$

Divide both sides by x^{2}.

$$\frac{x^{2}A}{x^{2}}=\frac{y-40}{x^{2}}$$

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

$$A=\frac{y-40}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-A^{-\frac{1}{2}}\sqrt{y-40}\text{; }x=A^{-\frac{1}{2}}\sqrt{y-40}\text{, }&A\neq 0\\x\in \mathrm{C}\text{, }&y=40\text{ and }A=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\sqrt{\frac{y-40}{A}}\text{; }x=-\sqrt{\frac{y-40}{A}}\text{, }&\left(y\geq 40\text{ and }A>0\right)\text{ or }\left(y\leq 40\text{ and }A<0\right)\\x\in \mathrm{R}\text{, }&y=40\text{ and }A=0\end{matrix}\right.$$