# Solve? 3n^2-2n+1=-3n^2+9n+11

$n = - \frac{2}{3} , \frac{5}{2}$

#### Explanation:

We have (I think I've got this right...)

$3 {n}^{2} - 2 n + 1 = - 3 {n}^{2} + 9 n + 11$

Let's set the whole thing equal to 0:

$6 {n}^{2} - 11 n - 10 = 0$

I'll try first to factor it into a form of $\left(a x + b\right) \left(c x + d\right)$ such that:

• $a c = 6$
• $b d = - 10$
• $a d + b c = - 11$

I'll do it in a chart format:

$\left(\begin{matrix}a & b & c & d & a c & b d & a d & b c & a d + b c \\ 3 & 2 & 2 & - 5 & 6 & - 10 & - 15 & 4 & - 11\end{matrix}\right)$

(first crack - got lucky!)

And so our factoring is:

$\left(3 n + 2\right) \left(2 n - 5\right) = 0$

$n = - \frac{2}{3} , \frac{5}{2}$

We can see these points of intersection in the graph:

graph{(y-(3x^2-2x+1))(y-(-3x^2+9x+11))=0 [-10, 10, -14.42, 20]}