Question 211092
First you plug in the 15 into the equation which gives you 15 = 2 + 28t - 5t^2.  Now all we have to do is solve for t.  To start I suggest we get everything on the same side of the equation.  <br>

15 = 2 + 28t - 5t^2
15-2-28t+5t^2 = 2-2 + 28t-28t-5t^2+5t^2
13-28t+5t^2 = 0<br>

Now lets just rearrange the equation so its in standard form.  
5t^2-28t+13 = 0<br>

Now try as we might this equation won't factor evenly so we have to use the quadratic equation.  <br>

{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} where a = 5, b = -28, and c = 13.  Plugging in those values gives us:<br>

{{{t = (28 +- sqrt( (-28)^2-4*5*13 ))/(2*5) }}}<br>

Now we just simplify:<br>

{{{t = (28 +- sqrt( (-28)^2-4*5*13 ))/(2*5) }}}
{{{t = (28 +- sqrt( 784-260 ))/(10) }}}
{{{t = (28 +- sqrt( 524 ))/(10) }}}
{{{t = (28 +- 2*sqrt( 131 ))/(10) }}}
{{{t = (14 +- sqrt( 131 ))/(5) }}}<br>

Plugging this into our calculator we get answers of 
t= 5.089104628 seconds or t = .510895 and I'm not sure which one we should use.