Question

Question #d15b1

Answer 1

see explanation

Explanation:

I'm going to interpret the request to "solve" the equations as "find the roots".
For quadratic equations, we can plug the coefficients into the well known forumula

`x = (-b +- sqrt(b^2 - 4ac))/(2a)`

w(t) is a quadratic equation, with coefficients -3, 0, and 5. The roots are 1.29 and -1.29

For the second part, I guess you can substitute (-2-t) in for the t parameter in the equation.

So, you'd have:

`-3(-2-t)^2 + 5` giving:

`-3(4 + 4t + t^2) + 5`

`= -12 -12t - 3t^2 + 5`

`= -3t^2 - 12t -7`

This is also a quadratic equation, with coefficients -3, -12, and -7. So you can solve for the roots, though we're not asked to do this. The roots are:

-3.29 and -0.71 (rounding)

GOOD LUCK