Question

Solve the following equation for v.

Answer 1

$b+\dfrac{v^2}{2d}=T\ \ \ |-b\\\\\dfrac{v^2}{2d}=T-b\ \ \ |\cdot2d\\\\v^2=2d(T-b)\to v=\pm\sqrt{2d(T-b)}$

Answer 2

Do the opposite of PEMDAS. First subtract b from both sides.

T  = b + v²/2d

T (-b) = b (-b) + v²/2d

T - b = v²/2d

Multiply 2d to both sides

2d(T - b) = v²/2d(2d)

2d(T - b) = v²

Distribute 2d to both T and -b

2dT - 2db = v²

Isolate the v, root both sides

√(2dT - 2db) = √v²

v = √(2dT - 2db) 

hope this helps