Solve 3y(8x+7y)y+9z=z^2

$$3 y ( 8 x + 7 y ) y + 9 z = z ^ { 2 }$$

Solve for x

$$\left\{\begin{matrix}x=\frac{-21y^{3}+z^{2}-9z}{24y^{2}}\text{, }&y\neq 0\\x\in \mathrm{R}\text{, }&\left(z=0\text{ or }z=9\right)\text{ and }y=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply y and y to get y^{2}.

$$3y^{2}\left(8x+7y\right)+9z=z^{2}$$

Use the distributive property to multiply 3y^{2} by 8x+7y.

$$24xy^{2}+21y^{3}+9z=z^{2}$$

Subtract 21y^{3} from both sides.

$$24xy^{2}+9z=z^{2}-21y^{3}$$

Subtract 9z from both sides.

$$24xy^{2}=z^{2}-21y^{3}-9z$$

The equation is in standard form.

$$24y^{2}x=-21y^{3}+z^{2}-9z$$

Divide both sides by 24y^{2}.

$$\frac{24y^{2}x}{24y^{2}}=\frac{-21y^{3}+z^{2}-9z}{24y^{2}}$$

Dividing by 24y^{2} undoes the multiplication by 24y^{2}.

$$x=\frac{-21y^{3}+z^{2}-9z}{24y^{2}}$$