Solve for n n(n+1)+3=45

$n (n + 1) + 3 = 45$

Step 1

Simplify each term.

Apply the distributive property.

$n \cdot n + n \cdot 1 + 3 = 45$

Multiply $n$ by $n$ .

$n^{2} + n \cdot 1 + 3 = 45$

Multiply $n$ by $1$ .

$n^{2} + n + 3 = 45$

Step 2

Subtract $45$ from both sides of the equation.

$n^{2} + n + 3 - 45 = 0$

Step 3

Subtract $45$ from $3$ .

$n^{2} + n - 42 = 0$

Step 4

Factor $n^{2} + n - 42$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 42$ and whose sum is $1$ .

$- 6, 7$

Write the factored form using these integers.

$(n - 6) (n + 7) = 0$

Step 5

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$n - 6 = 0$

$n + 7 = 0$

Step 6

Set $n - 6$ equal to $0$ and solve for $n$ .

Set $n - 6$ equal to $0$ .

$n - 6 = 0$

Add $6$ to both sides of the equation.

$n = 6$

Step 7

Set $n + 7$ equal to $0$ and solve for $n$ .

Set $n + 7$ equal to $0$ .

$n + 7 = 0$

Subtract $7$ from both sides of the equation.

$n = - 7$

Step 8

The final solution is all the values that make $(n - 6) (n + 7) = 0$ true.

$n = 6, - 7$

Solve for n n(n+1)+3=45

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