Find All Complex Solutions n^2+5n+1=3n

$n^{2} + 5 n + 1 = 3 n$

Move all terms to the left side of the equation and simplify.

Subtract $3 n$ from both sides of the equation.

$n^{2} + 5 n + 1 - 3 n = 0$

Subtract $3 n$ from $5 n$ .

$n^{2} + 2 n + 1 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 2$ , and $c = 1$ into the quadratic formula and solve for $n$ .

$\frac{- 2 \pm \sqrt{2^{2} - 4 \cdot (1 \cdot 1)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

Raise $2$ to the power of $2$ .

$n = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$n = \frac{- 2 \pm \sqrt{4 - 4 \cdot 1}}{2 \cdot 1}$

Multiply $- 4$ by $1$ .

$n = \frac{- 2 \pm \sqrt{4 - 4}}{2 \cdot 1}$

Subtract $4$ from $4$ .

$n = \frac{- 2 \pm \sqrt{0}}{2 \cdot 1}$

Rewrite $0$ as $0^{2}$ .

$n = \frac{- 2 \pm \sqrt{0^{2}}}{2 \cdot 1}$

Pull terms out from under the radical, assuming positive real numbers.

$n = \frac{- 2 \pm 0}{2 \cdot 1}$

$- 2$ plus or minus $0$ is $- 2$ .

$n = \frac{- 2}{2 \cdot 1}$

Multiply $2$ by $1$ .

$n = \frac{- 2}{2}$

Divide $- 2$ by $2$ .

$n = - 1$

The final answer is the combination of both solutions.

$n = - 1$ Double roots