Question 203347
Your problem is to find the value(s) of t that make h(t) = 119:
{{{119 = -16t^2 + 137t}}}
Subtract 119 from both sides:
{{{0 = -16t^2 + 137t - 119}}}
Factoring looks difficult at best and it may not factor. So we'll use the quadratic formula: {{{x = (-b +- sqrt(b^2 - 4ac))/(2a)}}} where a, b, and c are taken from the standard form for quadratic equations: ax^2 + bx + c. For your equation a = -16, b = 137 and c = -119:
{{{x = (-(137) +- sqrt(137^2 - 4(-16)(-119)))/(2(-16)) = (-137 +- sqrt(18769 - 4(-16)(-119)))/(-32) = (-137 +- sqrt(18769 - 7616))/(-32) = (-137 +- sqrt(11153))/(-32)}}}
Splitting this into the two solutions:
{{{x = (-137 + sqrt(11153))/(-32)}}} or {{{x = (-137 - sqrt(11153))/(-32)}}}