Answer:
[tex]n=\frac{3}{2},\,\,n=2[/tex].
Step-by-step explanation:
The equation you have is a quadratic equation because the polynomial [tex]2n^{2}-7n+6[/tex] has degree 2. One of the methods available to solve kind of equations is to factorize the polynomial on the left hand side. To factorize you can do the following:
(1) [tex]2n^2-7n+6[/tex]. The given polinomial
(2) [tex]\frac{2\times(2n^2-7n+6)}{2}=\frac{(2n)^{2}-7(2n)+12}{2}[/tex]. Multiply and divide by 2, because it is the coeficient of [tex]n^{2}[/tex]
(3) [tex]\frac{(2n)^{2}-7(2n)+12}{2}=\frac{(2n-\_\_)(2n-\_\_)}{2}[/tex]. Separate the polynomial in two factors, each one with [tex]2n[/tex] as a first term. The sign in the first factor is equal to the sign in the second term of the polynomial, that is to say, [tex]-7n[/tex]. The sign in the second factor is the sign of the second term multiplied by the sign of the third term, that is to say [tex](-)\times(+)=(-)[/tex] . In the blanks you should select two numbers whose sum is 7 and whose product is 12. Those numbers must be 3 and 4.
(4)The polynomial factorized is [tex]\frac{(2n-4)(2n-3)}{2}[/tex]
(5)Use the common factor in the numerator to cancel the number 2 in the denominator to obtain [tex](n-2)(2n-3)[/tex]
Then the given equation can be written as:
[tex]{(2n-3)(n-2)=0[/tex]
The product of two expression equals zero if and only if one of the expression is zero. From here we have that
[tex]2n-3=0[/tex] or [tex]n-2=0[/tex]
From the first equality we obtain that [tex]n=\frac{3}{2}[/tex]. From the second equality we obtain that [tex]n=2[/tex].
