We have
[tex]h=2+9t-5t^2[/tex]in order to know the time when h is equal to 3 we have
[tex]3=2+9t-5t^2[/tex]then
[tex]-1+9t-5t^2=0[/tex]then we solve this equation using the general formula for second-degree equations
[tex]t_{1,2}_{}=\frac{-9\pm\sqrt[]{9^2-4(-5)(-1)}}{2(-1)}[/tex][tex]t_{1,2}=\frac{-9\pm\sqrt[]{61}}{2(-5)}[/tex][tex]t_1=0.118=0.12[/tex][tex]t_2=1.681=1.68[/tex]We have two results when h=3, t1=0.12 seconds, and t2=1.68 seconds