Solve $\frac{n(n+1)}{2}= 45$ I have this equation:
$\frac{n(n+1)}{2}= 45$
I started:
$$n^2+n= 90$$
What do I do next?
 A: $n^2+n=90$ would be the correct starting point as each side is multiplied by 2.
As for next steps:
$n^2+n-90=0$
$\iff n^2+10n-9n-90=0$
$\iff n(n+10)-9(n+10)=0 $
$\iff (n+10)(n-9)=0$
Thus, $n$ is either 9 or -10.
A: Your equation is
$n(n+1)=90$
$=1\times90$
$=2\times45$
$=3\times30$
$=5\times18$
$=6\times15$
$=9\times10$
so $n=9$  or by opposite $n=-10$.
A: I assume you're solving for n.
Firstly, you didn't complete that step correctly. 
Multiplying both sides by 2 would give $n^2 + n = 90$.
 Subtracting 90 from both sides, $n^2 + n - 90 = 0$. 
Factoring gives $(n+10)(n-9)=0$ 
Therefore, the solutions are $n=9,n=-10$.
A: $n^2+n=90$
We can write this as-
$n^2+n-90=0$
Now, it is in the quadratic equation form, .i.e., $ax^2+bx+c=0$
Factor out the resulting equation, break $n$ into two terms $10n-9n$
$$n^2+10n-9n-90=0$$
Take $n$ common from first two terms and $-9$ from last two terms-
$$n(n+10)-9(n+10)=0$$
$$(n-9)(9+10)=0$$
$n-9=0\;\implies n=9$
$n+10=0\;\implies n=-10$
A: @Abdallah has already given an answer that doesn't involve solving a quadratic. My answer (at least the second way) is along the same lines, but maybe slightly more "optimised".
First way. Observe that the LHS is the exact formula for the $n$th triangular number. If you already know the first few terms of the sequence, you'll immediately be able to come up with the solution $n=9$. Also, if you keep in mind that there is a "mirror" sequence to the triangular numbers for the negative integers, albeit shifted by one, you'll be able to deduce the second solution $n=-10$.
Second way. Let's say you're not all that familiar with the triangular number sequence. Now you can follow Abdallah's method. However, instead of starting at $1$, you should start at $9$ and work backwards. This is because $n^2 + n$ is close to $n$ for larger $n$. So knowing $n^2 < 90$, you can look at the highest square that isn't greater than $90$, and that's $81$. By taking its square root, you have your starting value. In this case, you hit paydirt with the first guess.
