1. ## Quadratic equation - help

Code:
h=16t^2+vt+s (Solve for t)
h-h=16t^2+vt+s-h
0=16t^2+vt+s-h
a=16,b=v,c=s-h
t= (-b±[sqrt](b^2-4ac))/2a
t=(-v±[sqrt]((v)^2-4(16)(s-h) ))/2(16)
t=(-v±[sqrt](v^2-4(16s-16h) ))/32
t=(-v±[sqrt](v^2-64s+64h))/32
t=(-v±v+8s+8h)/32
t=(-v±)/32 v/32+8s/32+8h/32
Is this correct??

2. Originally Posted by mcruz65
h=16t^2+vt+s (Solve for t)
h-h=16t^2+vt+s-h
0=16t^2+vt+s-h
a=16,b=v,c=s-h
t= (-b±[sqrt](b^2-4ac))/2a
t=(-v±[sqrt]((v)^2-4(16)(s-h) ))/2(16)
t=(-v±[sqrt](v^2-4(16s-16h) ))/32
t=(-v±[sqrt](v^2-64s+64h))/32 ok up to here
t=(-v±v+8s+8h)/32 this is incorrect

$\displaystyle \textcolor{red}{\sqrt{v^2-64s+64h} \ne v+8s+8h}$
...

3. Originally Posted by mcruz65
Code:
h=16t^2+vt+s (Solve for t)
h-h=16t^2+vt+s-h
0=16t^2+vt+s-h
a=16,b=v,c=s-h
t= (-b±[sqrt](b^2-4ac))/2a
t=(-v±[sqrt]((v)^2-4(16)(s-h) ))/2(16)
t=(-v±[sqrt](v^2-4(16s-16h) ))/32
t=(-v±[sqrt](v^2-64s+64h))/32
t=(-v±v+8s+8h)/32
t=(-v±)/32 v/32+8s/32+8h/32
Is this correct??
Hi mcruz,

you cannot take the $\displaystyle \sqrt{v^2-64s+64h}$ by taking the square root of each term (almost).

$\displaystyle t=\frac{-v\pm\sqrt{v^2-64(s-h)}}{32}$