What is the answer to solvefor v,d=vt+1/2 at^2 ?

The answer to solvefor v,d=vt+1/2 at^2 is v=(2d-t^2a)/(2t);\quad t\ne 0

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solvefor v,d=vt+1/2 at^2

solve for

v, d = v t + \frac{1}{2} a t^{2}

v = \frac{2 d - t ^{2} a}{2 t}; t \neq = 0

Solution steps

d = v t + \frac{1}{2} a t^{2}

Switch sides

v t + \frac{1}{2} a t^{2} = d

Move

\frac{1}{2} a t^{2}

to the right side

v t = d - \frac{1}{2} a t^{2}

Divide both sides by

t; t \neq = 0

v = \frac{2 d - t ^{2} a}{2 t}; t \neq = 0

7.7 - 0.15 g = 0.1 g + 4.9

3 (x - 4) + 4 = 2 (x - 1)

- 2.9 = 3 f + 4.3

a, 2 a - 3 b = 5

x, 8 x - 3 y = 19

What is the answer to solvefor v,d=vt+1/2 at^2 ?
The answer to solvefor v,d=vt+1/2 at^2 is v=(2d-t^2a)/(2t);\quad t\ne 0