Question 171359
First, we need to factor {{{3t^2+11t-20}}}





Looking at {{{3t^2+11t-20}}} we can see that the first term is {{{3t^2}}} and the last term is {{{-20}}} where the coefficients are 3 and -20 respectively.


Now multiply the first coefficient 3 and the last coefficient -20 to get -60. Now what two numbers multiply to -60 and add to the  middle coefficient 11? Let's list all of the factors of -60:




Factors of -60:

1,2,3,4,5,6,10,12,15,20,30,60


-1,-2,-3,-4,-5,-6,-10,-12,-15,-20,-30,-60 ...List the negative factors as well. This will allow us to find all possible combinations


These factors pair up and multiply to -60

(1)*(-60)

(2)*(-30)

(3)*(-20)

(4)*(-15)

(5)*(-12)

(6)*(-10)

(-1)*(60)

(-2)*(30)

(-3)*(20)

(-4)*(15)

(-5)*(12)

(-6)*(10)


note: remember, the product of a negative and a positive number is a negative number



Now which of these pairs add to 11? Lets make a table of all of the pairs of factors we multiplied and see which two numbers add to 11


<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center">1</td><td align="center">-60</td><td>1+(-60)=-59</td></tr><tr><td align="center">2</td><td align="center">-30</td><td>2+(-30)=-28</td></tr><tr><td align="center">3</td><td align="center">-20</td><td>3+(-20)=-17</td></tr><tr><td align="center">4</td><td align="center">-15</td><td>4+(-15)=-11</td></tr><tr><td align="center">5</td><td align="center">-12</td><td>5+(-12)=-7</td></tr><tr><td align="center">6</td><td align="center">-10</td><td>6+(-10)=-4</td></tr><tr><td align="center">-1</td><td align="center">60</td><td>-1+60=59</td></tr><tr><td align="center">-2</td><td align="center">30</td><td>-2+30=28</td></tr><tr><td align="center">-3</td><td align="center">20</td><td>-3+20=17</td></tr><tr><td align="center">-4</td><td align="center">15</td><td>-4+15=11</td></tr><tr><td align="center">-5</td><td align="center">12</td><td>-5+12=7</td></tr><tr><td align="center">-6</td><td align="center">10</td><td>-6+10=4</td></tr></table>



From this list we can see that -4 and 15 add up to 11 and multiply to -60



Now looking at the expression {{{3t^2+11t-20}}}, replace {{{11t}}} with {{{-4t+15t}}} (notice {{{-4t+15t}}} adds up to {{{11t}}}. So it is equivalent to {{{11t}}})


{{{3t^2+highlight(-4t+15t)+-20}}}



Now let's factor {{{3t^2-4t+15t-20}}} by grouping:



{{{(3t^2-4t)+(15t-20)}}} Group like terms



{{{t(3t-4)+5(3t-4)}}} Factor out the GCF of {{{t}}} out of the first group. Factor out the GCF of {{{5}}} out of the second group



{{{(t+5)(3t-4)}}} Since we have a common term of {{{3t-4}}}, we can combine like terms


So {{{3t^2-4t+15t-20}}} factors to {{{(t+5)(3t-4)}}}



So this also means that {{{3t^2+11t-20}}} factors to {{{(t+5)(3t-4)}}} (since {{{3t^2+11t-20}}} is equivalent to {{{3t^2-4t+15t-20}}})



So {{{3t^2+11t-20}}} factors to {{{(t+5)(3t-4)}}}




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So this means that {{{3t^2+11t-20=0}}} is the same as {{{(t+5)(3t-4)=0}}}




{{{(t+5)(3t-4)=0}}} Start with the previous equation.



Now set each factor equal to zero:


{{{t+5=0}}} or  {{{3t-4=0}}} 



{{{t=-5}}} or  {{{t=4/3}}}    Now solve for t in each case



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Answer:



So the solutions are 


 {{{t=-5}}} or  {{{t=4/3}}}