Make #2t^(2) + 7t + 3# equal to zero to form a quadratic equation.
Here's the general form of the quadratic equation
#ax^(2) + bx + c =0#
You need to find two numbers that multiply to give #a * c# and that add to give #b#. For this example, the two numbers you need must multply to give #2 * 3 = 6# and add to give #"7"#.
To find them, just list all of the factors of #6# and try to find a pair that satisfies the criteria given
#6 = 1* 2 * 3#
In simple cases such as this one, these two numbers can be easily found to be #"6"# and #"1"#, since
#6 + 1 = 7# and #6 * 1 = 6#
So, your equation then becomes
#2t^(2) + 6t + t +3#
This then becomes
#(2t^(2) + 6t) + (t + 3) => 2t * (t + 3) + (t + 3)#, which can be written as
#(t + 3) * (2t + 1)# #-># this is how the original equation factors.
