Intermediate Algebra for College Students (7th Edition)

$v=\dfrac{-at^2+2s}{2t}$
Subtract $\frac{1}{2}at^2$ from both sides to find: $s-\frac{1}{2}at^2=\frac{1}{2}at^2 +vt - \frac{1}{2}at^2 \\s-\frac{1}{2}at^2=vt$ Divide both sides by $t$ to find: $\dfrac{s-\frac{1}{2}at^2}{t}=\dfrac{vt}{t} \\\dfrac{s-\frac{1}{2}at^2}{t}=v \\\dfrac{\frac{2s}{2} - \frac{at^2}{2}}{t}=v \\\dfrac{\frac{2s-at^2}{2}}{t}=v \\\dfrac{2s-at^2}{2t}=v \\v=\dfrac{-at^2+2s}{2t}$