H(t)=-16t^2+65t+3 how many seconds to reach max height. nearest tenth

Two ways to
A. Using calculus:
H(t)=-16t^2+65t+3
To find maximum of H(t), equate H'(t)=0, or
H'(t)=-32t+65=0 => t=65/32=2.03125, or
t=2.0 s. to the nearest 10

B, Without using calculus, i.e. by completing the square
H(t)
=-16t^2+65t+3
=-16(t^2-65/16t)+3
=-16(t^2-2*(65/32)t+(65/32)^2)+3+(65/32)^2
=-16(t-65/32)^2+3+(65/32)^2
meaning the maximum occurs at (t-65/32)=0, or t=65/32=2.03125
t=2.0 seconds (to the nearest tenth)