Question 307569
Solve for t when s = 19:
{{{s = 2t^2-3t}}} Substitute s = 19.
{{{19 = 2t^2-3t}}} Rearrange into standard quadratic equation form:
{{{2t^2-3t-19 = 0}}} Solve using the quadratic formula:{{{t = (-b+-sqrt(b^2-4ac))/2a}}} where a = 2, b = -3, and c = -19.
{{{t = (-(-3)+-sqrt((-3)^2-4(2)(-19)))/2(2)}}}
{{{t = (3+-sqrt(9+152))/4}}}
{{{t = (3+-sqrt(161))/4}}}
{{{t = 3.9}}} or {{{t = -2.4}}} Discard the negative solution as the time, t, should be a positive quantity.
{{{highlight(t = 3.9)}}}seconds.