Question 132818
{{{17=2t^2-3t}}} Start with the given equation



{{{17=2t^2-3t}}} Plug in {{{s=17}}}



{{{0=2t^2-3t-17}}}  Subtract 17 from both sides. 


Let's use the quadratic formula to solve for t:



Starting with the general quadratic


{{{at^2+bt+c=0}}}


the general solution using the quadratic equation is:


{{{t = (-b +- sqrt( b^2-4*a*c ))/(2*a)}}}




So lets solve {{{0=2*t^2-3*t-17}}} ( notice {{{a=2}}}, {{{b=-3}}}, and {{{c=-17}}})





{{{t = (--3 +- sqrt( (-3)^2-4*2*-17 ))/(2*2)}}} Plug in a=2, b=-3, and c=-17




{{{t = (3 +- sqrt( (-3)^2-4*2*-17 ))/(2*2)}}} Negate -3 to get 3




{{{t = (3 +- sqrt( 9-4*2*-17 ))/(2*2)}}} Square -3 to get 9  (note: remember when you square -3, you must square the negative as well. This is because {{{(-3)^2=-3*-3=9}}}.)




{{{t = (3 +- sqrt( 9+136 ))/(2*2)}}} Multiply {{{-4*-17*2}}} to get {{{136}}}




{{{t = (3 +- sqrt( 145 ))/(2*2)}}} Combine like terms in the radicand (everything under the square root)





{{{t = (3 +- sqrt(145))/4}}} Multiply 2 and 2 to get 4


So now the expression breaks down into two parts


{{{t = (3 + sqrt(145))/4}}} or {{{t = (3 - sqrt(145))/4}}}



So these expressions approximate to


{{{t=3.76039864469807}}} or {{{t=-2.26039864469807}}}



So our possible solutions are:


{{{t=3.76039864469807}}} or {{{t=-2.26039864469807}}}




However, since a negative time doesn't make sense, our only solution is {{{t=3.76039864469807}}}



So it takes about 3.76 seconds for the object to move 17 meters