Question 149304
{{{5t^2-16t=-12}}} Start with the given equation.



{{{5t^2-16t+12=0}}} Add 12 to both sides.



Now let's factor the left side







Looking at the expression {{{5t^2-16t+12}}}, we can see that the first coefficient is {{{5}}}, the second coefficient is {{{-16}}}, and the last term is {{{12}}}.



Now multiply the first coefficient {{{5}}} by the last term {{{12}}} to get {{{(5)(12)=60}}}.



Now the question is: what two whole numbers multiply to {{{60}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-16}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{60}}} (the previous product).



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



Note: list the negative of each factor. 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)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-16}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>60</font></td><td  align="center"><font color=black>1+60=61</font></td></tr><tr><td  align="center"><font color=black>2</font></td><td  align="center"><font color=black>30</font></td><td  align="center"><font color=black>2+30=32</font></td></tr><tr><td  align="center"><font color=black>3</font></td><td  align="center"><font color=black>20</font></td><td  align="center"><font color=black>3+20=23</font></td></tr><tr><td  align="center"><font color=black>4</font></td><td  align="center"><font color=black>15</font></td><td  align="center"><font color=black>4+15=19</font></td></tr><tr><td  align="center"><font color=black>5</font></td><td  align="center"><font color=black>12</font></td><td  align="center"><font color=black>5+12=17</font></td></tr><tr><td  align="center"><font color=black>6</font></td><td  align="center"><font color=black>10</font></td><td  align="center"><font color=black>6+10=16</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-60</font></td><td  align="center"><font color=black>-1+(-60)=-61</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-30</font></td><td  align="center"><font color=black>-2+(-30)=-32</font></td></tr><tr><td  align="center"><font color=black>-3</font></td><td  align="center"><font color=black>-20</font></td><td  align="center"><font color=black>-3+(-20)=-23</font></td></tr><tr><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-15</font></td><td  align="center"><font color=black>-4+(-15)=-19</font></td></tr><tr><td  align="center"><font color=black>-5</font></td><td  align="center"><font color=black>-12</font></td><td  align="center"><font color=black>-5+(-12)=-17</font></td></tr><tr><td  align="center"><font color=red>-6</font></td><td  align="center"><font color=red>-10</font></td><td  align="center"><font color=red>-6+(-10)=-16</font></td></tr></table>



From the table, we can see that the two numbers {{{-6}}} and {{{-10}}} add to {{{-16}}} (the middle coefficient).



So the two numbers {{{-6}}} and {{{-10}}} both multiply to {{{60}}} <font size=4><b>and</b></font> add to {{{-16}}}



Now replace the middle term {{{-16t}}} with {{{-6t-10t}}}. Remember, {{{-6}}} and {{{-10}}} add to {{{-16}}}. So this shows us that {{{-6t-10t=-16t}}}.



{{{5t^2+highlight(-6t-10t)+12}}} Replace the second term {{{-16t}}} with {{{-6t-10t}}}.



{{{(5t^2-6t)+(-10t+12)}}} Group the terms into two pairs.



{{{t(5t-6)+(-10t+12)}}} Factor out the GCF {{{t}}} from the first group.



{{{t(5t-6)-2(5t-6)}}} Factor out {{{2}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(t-2)(5t-6)}}} Combine like terms. Or factor out the common term {{{5t-6}}}




So {{{5t^2-16t+12}}} factors to {{{(t-2)(5t-6)}}}.



Note: you can check to see if you did it right by FOILing {{{(t-2)(5t-6)}}} to get {{{5t^2-16t+12}}}.




So {{{5t^2-16t+12=0}}} becomes  {{{(t-2)(5t-6)=0}}}.




Now set each factor equal to zero:



{{{t-2=0}}} or  {{{5t-6=0}}} 


{{{t=2}}} or  {{{t=6/5}}}    Now solve for t in each case



So our answers are 


{{{t=2}}} or  {{{t=6/5}}}