Let:
[tex]h(t)=-5t^2+14t+3[/tex]We need to know when:
[tex]h(t)=0[/tex]so:
[tex]\begin{gathered} -5t^2+14t+3=0 \\ -(5t^2-14t-3)=0 \end{gathered}[/tex]The coefficient of t² is 5 and the constant term is -3. The product of 5 and -3 is -15, the factors of -15 which sum to -14 are 1 and -15. so:
[tex]\begin{gathered} 5t^2-14t-3=(t-3)(5t+1) \\ so\colon \\ -(5t^2-14t-3)=-(t-3)(5t+1) \end{gathered}[/tex]Therefore:
[tex]\begin{gathered} t=-\frac{1}{5} \\ t=3 \end{gathered}[/tex]Since -1/5 s wouldn't make any sense, the answer is:
t = 3 seconds
