Solve [A]quady=(x+7)^2-47

$$[ A ] \quad y = ( x + 7 ) ^ { 2 } - 47$$

Solve for A (complex solution)

$$\left\{\begin{matrix}A=\frac{x^{2}+14x+2}{y}\text{, }&y\neq 0\\A\in \mathrm{C}\text{, }&\left(x=\sqrt{47}-7\text{ or }x=-\left(\sqrt{47}+7\right)\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$Ay=x^{2}+14x+49-47$$

Subtract 47 from 49 to get 2.

$$Ay=x^{2}+14x+2$$

The equation is in standard form.

$$yA=x^{2}+14x+2$$

Divide both sides by y.

$$\frac{yA}{y}=\frac{x^{2}+14x+2}{y}$$

Dividing by y undoes the multiplication by y.

$$A=\frac{x^{2}+14x+2}{y}$$

Solve for A

$$\left\{\begin{matrix}A=\frac{x^{2}+14x+2}{y}\text{, }&y\neq 0\\A\in \mathrm{R}\text{, }&\left(x=-\left(\sqrt{47}+7\right)\text{ or }x=\sqrt{47}-7\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$Ay=x^{2}+14x+49-47$$

Subtract 47 from 49 to get 2.

$$Ay=x^{2}+14x+2$$

The equation is in standard form.

$$yA=x^{2}+14x+2$$

Divide both sides by y.

$$\frac{yA}{y}=\frac{x^{2}+14x+2}{y}$$

Dividing by y undoes the multiplication by y.

$$A=\frac{x^{2}+14x+2}{y}$$

Solve for x (complex solution)

$$x=-\left(\sqrt{Ay+47}+7\right)$$

$$x=\sqrt{Ay+47}-7$$

Solve for x

$$\left\{\begin{matrix}x=-7\text{, }&A=-\frac{47}{y}\text{ and }y\neq 0\\x=-\left(\sqrt{Ay+47}+7\right)\text{; }x=\sqrt{Ay+47}-7\text{, }&\left(y<0\text{ and }A\leq -\frac{47}{y}\right)\text{ or }\left(y>0\text{ and }A\geq -\frac{47}{y}\right)\\x=-\left(\sqrt{47}+7\right)\text{; }x=\sqrt{47}-7\text{, }&y=0\end{matrix}\right.$$