Solve (5k-b)(t+2b)=48

$$( 5 k - b ) ( t + 2 b ) = 48$$

Solve for k

$$k=\frac{bt+2b^{2}+48}{5\left(t+2b\right)}$$

$$b\neq -\frac{t}{2}$$

Steps for Solving Linear Equation

Use the distributive property to multiply 5k-b by t+2b.

$$5kt+10kb-bt-2b^{2}=48$$

Add bt to both sides.

$$5kt+10kb-2b^{2}=48+bt$$

Add 2b^{2} to both sides.

$$5kt+10kb=48+bt+2b^{2}$$

Combine all terms containing k.

$$\left(5t+10b\right)k=48+bt+2b^{2}$$

The equation is in standard form.

$$\left(5t+10b\right)k=bt+2b^{2}+48$$

Divide both sides by 5t+10b.

$$\frac{\left(5t+10b\right)k}{5t+10b}=\frac{bt+2b^{2}+48}{5t+10b}$$

Dividing by 5t+10b undoes the multiplication by 5t+10b.

$$k=\frac{bt+2b^{2}+48}{5t+10b}$$

Divide 48+bt+2b^{2} by 5t+10b.

$$k=\frac{bt+2b^{2}+48}{5\left(t+2b\right)}$$

Solve for b (complex solution)

$$b=-\frac{\sqrt{t^{2}+20kt+100k^{2}-384}}{4}+\frac{5k}{2}-\frac{t}{4}$$

$$b=\frac{\sqrt{t^{2}+20kt+100k^{2}-384}}{4}+\frac{5k}{2}-\frac{t}{4}$$

Solve for b

$$b=-\frac{\sqrt{t^{2}+20kt+100k^{2}-384}}{4}+\frac{5k}{2}-\frac{t}{4}$$

$$b=\frac{\sqrt{t^{2}+20kt+100k^{2}-384}}{4}+\frac{5k}{2}-\frac{t}{4}\text{, }k\leq -\frac{t}{10}-\frac{4\sqrt{6}}{5}\text{ or }k\geq -\frac{t}{10}+\frac{4\sqrt{6}}{5}$$