a baseball is thrown into the air with an upward velocity of 30 ft/s. its initial height was 6 ft, and its maximum height is 20.06 ft. how long will it take the ball to reach its maximum height? round to the nearest hundredth.i've been stuck on this question for almost an hour so if anyone can help that would be greatly appreciated

Check the picture below.where is the -16t² coming from?  that's Earth's gravity pull in feet.[tex]\bf ~~~~~~\textit{initial velocity} \\\\ \begin{array}{llll} ~~~~~~\textit{in feet} \\\\ h(t) = -16t^2+v_ot+h_o \end{array} \quad \begin{cases} v_o=\stackrel{30}{\textit{initial velocity of the object}}\\\\ h_o=\stackrel{6}{\textit{initial height of the object}}\\\\ h=\stackrel{}{\textit{height of the object at "t" seconds}} \end{cases} \\\\\\ h(t)=-16t^2+30t+6 \\\\[-0.35em] ~\dotfill[/tex][tex]\bf \textit{vertex of a vertical parabola, using coefficients} \\\\ h(t)=\stackrel{\stackrel{a}{\downarrow }}{-16}t^2\stackrel{\stackrel{b}{\downarrow }}{+30}t\stackrel{\stackrel{c}{\downarrow }}{+6} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)[/tex][tex]\bf \left(-\cfrac{30}{2(-16)}~~,~~6-\cfrac{30^2}{4(-16)} \right)\implies \left( \cfrac{30}{32}~,~6+\cfrac{225}{16} \right)\implies \left(\cfrac{15}{16}~,~\cfrac{321}{16} \right) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill (\stackrel{\stackrel{\textit{how many}}{\textit{seconds it took}}}{0.9375}~~,~~\stackrel{\stackrel{\textit{how many feet}}{\textit{up it went}}}{20.0625})~\hfill[/tex]