Find All Complex Solutions 4n^2+24n+25=4n

$4 n^{2} + 24 n + 25 = 4 n$

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

Subtract $4 n$ from both sides of the equation.

$4 n^{2} + 24 n + 25 - 4 n = 0$

Subtract $4 n$ from $24 n$ .

$4 n^{2} + 20 n + 25 = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 4$ , $b = 20$ , and $c = 25$ into the quadratic formula and solve for $n$ .

$\frac{- 20 \pm \sqrt{20^{2} - 4 \cdot (4 \cdot 25)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $20$ to the power of $2$ .

$n = \frac{- 20 \pm \sqrt{400 - 4 \cdot 4 \cdot 25}}{2 \cdot 4}$

Multiply $- 4$ by $4$ .

$n = \frac{- 20 \pm \sqrt{400 - 16 \cdot 25}}{2 \cdot 4}$

Multiply $- 16$ by $25$ .

$n = \frac{- 20 \pm \sqrt{400 - 400}}{2 \cdot 4}$

Subtract $400$ from $400$ .

$n = \frac{- 20 \pm \sqrt{0}}{2 \cdot 4}$

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

$n = \frac{- 20 \pm \sqrt{0^{2}}}{2 \cdot 4}$

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

$n = \frac{- 20 \pm 0}{2 \cdot 4}$

$- 20$ plus or minus $0$ is $- 20$ .

$n = \frac{- 20}{2 \cdot 4}$

Multiply $2$ by $4$ .

$n = \frac{- 20}{8}$

Cancel the common factor of $- 20$ and $8$ .

Factor $4$ out of $- 20$ .

$n = \frac{4 (- 5)}{8}$

Cancel the common factors.

Factor $4$ out of $8$ .

$n = \frac{4 \cdot - 5}{4 \cdot 2}$

Cancel the common factor.

$n = \frac{4 \cdot - 5}{4 \cdot 2}$

Rewrite the expression.

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

Move the negative in front of the fraction.

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

The final answer is the combination of both solutions.

$n = - \frac{5}{2}$ Double roots