Respuesta :

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]

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