Solve ty^2-48y+18=9

$$t y ^ { 2 } - 48 y + 18 = 9$$

Solve for t

$$t=-\frac{3\left(3-16y\right)}{y^{2}}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Add 48y to both sides.

$$ty^{2}+18=9+48y$$

Subtract 18 from both sides.

$$ty^{2}=9+48y-18$$

Subtract 18 from 9 to get -9.

$$ty^{2}=-9+48y$$

The equation is in standard form.

$$y^{2}t=48y-9$$

Divide both sides by y^{2}.

$$\frac{y^{2}t}{y^{2}}=\frac{48y-9}{y^{2}}$$

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

$$t=\frac{48y-9}{y^{2}}$$

Divide -9+48y by y^{2}.

$$t=\frac{3\left(16y-3\right)}{y^{2}}$$

Solve for y (complex solution)

$$\left\{\begin{matrix}y=\frac{3\left(\sqrt{64-t}+8\right)}{t}\text{; }y=\frac{3\left(-\sqrt{64-t}+8\right)}{t}\text{, }&t\neq 0\\y=\frac{3}{16}\text{, }&t=0\end{matrix}\right.$$

Solve for y

$$\left\{\begin{matrix}y=\frac{3\left(\sqrt{64-t}+8\right)}{t}\text{; }y=\frac{3\left(-\sqrt{64-t}+8\right)}{t}\text{, }&t\neq 0\text{ and }t\leq 64\\y=\frac{3}{16}\text{, }&t=0\end{matrix}\right.$$