Answer:
The answer is B.
[tex]t=+\sqrt{\frac{d-3}{a} } , -\sqrt{\frac{d-3}{a}[/tex]
Step-by-step explanation:
[tex]d=3+at^2\\\\d-3=at^2\\\\\frac{d-3}{a}=t^2\\\\ t=+\sqrt{\frac{d-3}{a} } , -\sqrt{\frac{d-3}{a} }[/tex]
