Solve (a_{3}+8)^2=(a_{2}+10)(a_{4}+6)

$$( a _ { 3 } + 8 ) ^ { 2 } = ( a _ { 2 } + 10 ) ( a _ { 4 } + 6 )$$

Solve for a_2 (complex solution)

$$\left\{\begin{matrix}a_{2}=-\frac{-a_{3}^{2}-16a_{3}+10a_{4}-4}{a_{4}+6}\text{, }&a_{4}\neq -6\\a_{2}\in \mathrm{C}\text{, }&a_{3}=-8\text{ and }a_{4}=-6\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a_{3}+8\right)^{2}.

$$a_{3}^{2}+16a_{3}+64=\left(a_{2}+10\right)\left(a_{4}+6\right)$$

Use the distributive property to multiply a_{2}+10 by a_{4}+6.

$$a_{3}^{2}+16a_{3}+64=a_{2}a_{4}+6a_{2}+10a_{4}+60$$

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

$$a_{2}a_{4}+6a_{2}+10a_{4}+60=a_{3}^{2}+16a_{3}+64$$

Subtract 10a_{4} from both sides.

$$a_{2}a_{4}+6a_{2}+60=a_{3}^{2}+16a_{3}+64-10a_{4}$$

Subtract 60 from both sides.

$$a_{2}a_{4}+6a_{2}=a_{3}^{2}+16a_{3}+64-10a_{4}-60$$

Subtract 60 from 64 to get 4.

$$a_{2}a_{4}+6a_{2}=a_{3}^{2}+16a_{3}+4-10a_{4}$$

Combine all terms containing a_{2}.

$$\left(a_{4}+6\right)a_{2}=a_{3}^{2}+16a_{3}+4-10a_{4}$$

The equation is in standard form.

$$\left(a_{4}+6\right)a_{2}=a_{3}^{2}+16a_{3}-10a_{4}+4$$

Divide both sides by a_{4}+6.

$$\frac{\left(a_{4}+6\right)a_{2}}{a_{4}+6}=\frac{a_{3}^{2}+16a_{3}-10a_{4}+4}{a_{4}+6}$$

Dividing by a_{4}+6 undoes the multiplication by a_{4}+6.

$$a_{2}=\frac{a_{3}^{2}+16a_{3}-10a_{4}+4}{a_{4}+6}$$

Solve for a_2

$$\left\{\begin{matrix}a_{2}=-\frac{-a_{3}^{2}-16a_{3}+10a_{4}-4}{a_{4}+6}\text{, }&a_{4}\neq -6\\a_{2}\in \mathrm{R}\text{, }&a_{3}=-8\text{ and }a_{4}=-6\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(a_{3}+8\right)^{2}.

$$a_{3}^{2}+16a_{3}+64=\left(a_{2}+10\right)\left(a_{4}+6\right)$$

Use the distributive property to multiply a_{2}+10 by a_{4}+6.

$$a_{3}^{2}+16a_{3}+64=a_{2}a_{4}+6a_{2}+10a_{4}+60$$

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

$$a_{2}a_{4}+6a_{2}+10a_{4}+60=a_{3}^{2}+16a_{3}+64$$

Subtract 10a_{4} from both sides.

$$a_{2}a_{4}+6a_{2}+60=a_{3}^{2}+16a_{3}+64-10a_{4}$$

Subtract 60 from both sides.

$$a_{2}a_{4}+6a_{2}=a_{3}^{2}+16a_{3}+64-10a_{4}-60$$

Subtract 60 from 64 to get 4.

$$a_{2}a_{4}+6a_{2}=a_{3}^{2}+16a_{3}+4-10a_{4}$$

Combine all terms containing a_{2}.

$$\left(a_{4}+6\right)a_{2}=a_{3}^{2}+16a_{3}+4-10a_{4}$$

The equation is in standard form.

$$\left(a_{4}+6\right)a_{2}=a_{3}^{2}+16a_{3}-10a_{4}+4$$

Divide both sides by a_{4}+6.

$$\frac{\left(a_{4}+6\right)a_{2}}{a_{4}+6}=\frac{a_{3}^{2}+16a_{3}-10a_{4}+4}{a_{4}+6}$$

Dividing by a_{4}+6 undoes the multiplication by a_{4}+6.

$$a_{2}=\frac{a_{3}^{2}+16a_{3}-10a_{4}+4}{a_{4}+6}$$

Solve for a_3 (complex solution)

$$a_{3}=-\left(\sqrt{a_{2}+10}\sqrt{a_{4}+6}+8\right)$$

$$a_{3}=\sqrt{a_{2}+10}\sqrt{a_{4}+6}-8$$

Solve for a_3

$$a_{3}=-\left(\sqrt{\left(a_{2}+10\right)\left(a_{4}+6\right)}+8\right)$$

$$a_{3}=\sqrt{\left(a_{2}+10\right)\left(a_{4}+6\right)}-8\text{, }\left(a_{2}\geq -10\text{ and }a_{4}\geq -6\right)\text{ or }\left(a_{4}\leq -6\text{ and }a_{2}\leq -10\right)$$