Solve 0.01m+0.2=-1/160(t-24)^2+0.4

$$0.01 m + 0.2 = - \frac { 1 } { 160 } ( t - 24 ) ^ { 2 } + 0.4$$

Solve for m

$$m=-\frac{5t^{2}}{8}+30t-340$$

Steps for Solving Linear Equation

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(t-24\right)^{2}.

$$0.01m+0.2=-\frac{1}{160}\left(t^{2}-48t+576\right)+0.4$$

Use the distributive property to multiply -\frac{1}{160} by t^{2}-48t+576.

$$0.01m+0.2=-\frac{1}{160}t^{2}+\frac{3}{10}t-\frac{18}{5}+0.4$$

Add -\frac{18}{5} and 0.4 to get -\frac{16}{5}.

$$0.01m+0.2=-\frac{1}{160}t^{2}+\frac{3}{10}t-\frac{16}{5}$$

Subtract 0.2 from both sides.

$$0.01m=-\frac{1}{160}t^{2}+\frac{3}{10}t-\frac{16}{5}-0.2$$

Subtract 0.2 from -\frac{16}{5} to get -\frac{17}{5}.

$$0.01m=-\frac{1}{160}t^{2}+\frac{3}{10}t-\frac{17}{5}$$

The equation is in standard form.

$$0.01m=-\frac{t^{2}}{160}+\frac{3t}{10}-\frac{17}{5}$$

Multiply both sides by 100.

$$\frac{0.01m}{0.01}=\frac{-\frac{t^{2}}{160}+\frac{3t}{10}-\frac{17}{5}}{0.01}$$

Dividing by 0.01 undoes the multiplication by 0.01.

$$m=\frac{-\frac{t^{2}}{160}+\frac{3t}{10}-\frac{17}{5}}{0.01}$$

Divide -\frac{t^{2}}{160}+\frac{3t}{10}-\frac{17}{5} by 0.01 by multiplying -\frac{t^{2}}{160}+\frac{3t}{10}-\frac{17}{5} by the reciprocal of 0.01.

$$m=-\frac{5t^{2}}{8}+30t-340$$

Solve for t

$$t=-\frac{2\sqrt{200-10m}}{5}+24$$

$$t=\frac{2\sqrt{200-10m}}{5}+24\text{, }m\leq 20$$