Solve 2n^2=?4n

$$2 n ^ { 2 } = ? 4 n$$

Solve for n

$$n=2$$

$$n=0$$

Steps Using Factoring

Subtract 4n from both sides.

$$2n^{2}-4n=0$$

Factor out n.

$$n\left(2n-4\right)=0$$

To find equation solutions, solve n=0 and 2n-4=0.

$$n=0 n=2$$

Steps Using the Quadratic Formula

Subtract 4n from both sides.

$$2n^{2}-4n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -4 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}}}{2\times 2}$$

Take the square root of \left(-4\right)^{2}.

$$n=\frac{-\left(-4\right)±4}{2\times 2}$$

The opposite of -4 is 4.

$$n=\frac{4±4}{2\times 2}$$

Multiply 2 times 2.

$$n=\frac{4±4}{4}$$

Now solve the equation n=\frac{4±4}{4} when ± is plus. Add 4 to 4.

$$n=\frac{8}{4}$$

Divide 8 by 4.

$$n=2$$

Now solve the equation n=\frac{4±4}{4} when ± is minus. Subtract 4 from 4.

$$n=\frac{0}{4}$$

Divide 0 by 4.

$$n=0$$

The equation is now solved.

$$n=2 n=0$$

Steps for Completing the Square

Subtract 4n from both sides.

$$2n^{2}-4n=0$$

Divide both sides by 2.

$$\frac{2n^{2}-4n}{2}=\frac{0}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$n^{2}+\left(-\frac{4}{2}\right)n=\frac{0}{2}$$

Divide -4 by 2.

$$n^{2}-2n=\frac{0}{2}$$

Divide 0 by 2.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Factor n^{2}-2n+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-1\right)^{2}=1$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-1\right)^{2}}=\sqrt{1}$$

Simplify.

$$n-1=1 n-1=-1$$

Add 1 to both sides of the equation.

$$n=2 n=0$$