Solve for x x^2-4nx+4m>0

$x^{2} - 4 n x + 4 m > 0$

Convert the inequality to an equation.

$x^{2} - 4 n x + 4 m = 0$

Use the quadratic formula to find the solutions.

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

Substitute the values $a = 1$ , $b = - 4 n$ , and $c = 4 m$ into the quadratic formula and solve for $x$ .

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

Simplify.

Simplify the numerator.

Add parentheses.

$x = \frac{4 n \pm \sqrt{{(- 4 n)}^{2} - 4 \cdot (1 \cdot (4 m))}}{2 \cdot 1}$

Let $u = 1 \cdot (4 m)$ . Substitute $u$ for all occurrences of $1 \cdot (4 m)$ .

Apply the product rule to $- 4 n$ .

$x = \frac{4 n \pm \sqrt{{(- 4)}^{2} n^{2} - 4 \cdot u}}{2 \cdot 1}$

Raise $- 4$ to the power of $2$ .

$x = \frac{4 n \pm \sqrt{16 n^{2} - 4 u}}{2 \cdot 1}$

Factor $4$ out of $16 n^{2} - 4 u$ .

Factor $4$ out of $16 n^{2}$ .

$x = \frac{4 n \pm \sqrt{4 (4 n^{2}) - 4 u}}{2 \cdot 1}$

Factor $4$ out of $- 4 u$ .

$x = \frac{4 n \pm \sqrt{4 (4 n^{2}) + 4 (- u)}}{2 \cdot 1}$

Factor $4$ out of $4 (4 n^{2}) + 4 (- u)$ .

$x = \frac{4 n \pm \sqrt{4 (4 n^{2} - u)}}{2 \cdot 1}$

Replace all occurrences of $u$ with $1 \cdot (4 m)$ .

$x = \frac{4 n \pm \sqrt{4 (4 n^{2} - (1 \cdot (4 m)))}}{2 \cdot 1}$

Simplify each term.

Multiply $4 m$ by $1$ .

$x = \frac{4 n \pm \sqrt{4 (4 n^{2} - (4 m))}}{2 \cdot 1}$

Multiply $4$ by $- 1$ .

$x = \frac{4 n \pm \sqrt{4 (4 n^{2} - 4 m)}}{2 \cdot 1}$

Factor $4$ out of $4 n^{2} - 4 m$ .

Factor $4$ out of $4 n^{2}$ .

$x = \frac{4 n \pm \sqrt{4 (4 (n^{2}) - 4 m)}}{2 \cdot 1}$

Factor $4$ out of $- 4 m$ .

$x = \frac{4 n \pm \sqrt{4 (4 (n^{2}) + 4 (- m))}}{2 \cdot 1}$

Factor $4$ out of $4 (n^{2}) + 4 (- m)$ .

$x = \frac{4 n \pm \sqrt{4 (4 (n^{2} - m))}}{2 \cdot 1}$

$x = \frac{4 n \pm \sqrt{4 \cdot (4 (n^{2} - m))}}{2 \cdot 1}$

Multiply $4$ by $4$ .

$x = \frac{4 n \pm \sqrt{16 (n^{2} - m)}}{2 \cdot 1}$

Rewrite $16 (n^{2} - m)$ as ${(2^{2})}^{2} (n^{2} - m)$ .

Rewrite $16$ as $4^{2}$ .

$x = \frac{4 n \pm \sqrt{4^{2} (n^{2} - m)}}{2 \cdot 1}$

Rewrite $4$ as $2^{2}$ .

$x = \frac{4 n \pm \sqrt{{(2^{2})}^{2} (n^{2} - m)}}{2 \cdot 1}$

Pull terms out from under the radical.

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

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

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

Multiply $2$ by $1$ .

$x = \frac{4 n \pm 4 \sqrt{n^{2} - m}}{2}$

Simplify $\frac{4 n \pm 4 \sqrt{n^{2} - m}}{2}$ .

$x = 2 n \pm 2 \sqrt{n^{2} - m}$

Consolidate the solutions.

$x = 2 n + 2 \sqrt{n^{2} - m}$

$x = 2 n - 2 \sqrt{n^{2} - m}$

Solve for x x^2-4nx+4m>0

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