For this case we must factor the following expression:
[tex]3k ^ 2-19 = 15k-1[/tex]
We manipulate algebraically, taking into account that different signs are subtracted and the sign of the major is placed:
[tex]3k ^ 2-15k-19 + 1 = 0\\3k ^ 2-15k-18 = 0[/tex]
We divide by 3 on both sides of the equation:
[tex]k ^ 2-5k-6 = 0[/tex]
To factor, we look for two numbers that, when multiplied, result in -6 and when added, result in -5. These numbers are -6 and +1.
[tex]-6 + 1 = -5\\-6 * (+ 1) = - 6[/tex]
Thus, we factor the equation:
[tex](k-6) (k + 1) = 0[/tex]
The roots of the equation are:
[tex]k_ {1} = 6\\k_ {2} = - 1[/tex]
ANswer:
[tex]k_ {1} = 6\\k_ {2} = - 1[/tex]
