Question 34328
{{{ 4/(m^2-7) = 1 }}}
{{{ 4 = m^2-7 }}}
{{{ 11 = m^2 }}}
{{{ m = sqrt(11) }}} or {{{ m = - sqrt(11) }}}


{{{ m = 16t^2+vt+7 }}}
{{{ (m-7)/16 = t^2+(v/16)t }}}


let 2a = v/16
so a = v/32


therefore having {{{ (m-7)/16 = t^2+(v/16)t + (v/32)^2 - (v/32)^2 }}} is still the same thing as before, however we can now write it as:


{{{ (m-7)/16 = (t+(v/32))^2 - (v/32)^2 }}}
{{{ (m-7)/16 + (v/32)^2 = (t+(v/32))^2 }}}
{{{ (m-7)/16 + (v^2/1024) = (t+(v/32))^2 }}}
{{{ (64(m-7))/1024 + (v^2/1024) = (t+(v/32))^2 }}}
{{{ (64m-448)/1024 + (v^2/1024) = (t+(v/32))^2 }}}
{{{ (64m-448+v^2)/1024 = (t+(v/32))^2 }}}
+- {{{ sqrt((64m-448+v^2)/1024) = t+(v/32) }}}
{{{ (-v/32) +- sqrt((64m-448+v^2)/1024) = t }}}
{{{ (-v/32) +- (sqrt(64m-448+v^2))/32 = t }}}
{{{ (-v +- sqrt(64m-448+v^2))/32 = t }}}


jon.