Solve -5n(2n-1)=0

$$- 5 n ( 2 n - 1 ) = 0$$

Solve for n

$$n=\frac{1}{2}=0.5$$

$$n=0$$

Steps Using Factoring

Use the distributive property to multiply -5n by 2n-1.

$$-10n^{2}+5n=0$$

Factor out n.

$$n\left(-10n+5\right)=0$$

To find equation solutions, solve n=0 and -10n+5=0.

$$n=0 n=\frac{1}{2}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -5n by 2n-1.

$$-10n^{2}+5n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 5 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-5±\sqrt{5^{2}}}{2\left(-10\right)}$$

Take the square root of 5^{2}.

$$n=\frac{-5±5}{2\left(-10\right)}$$

Multiply 2 times -10.

$$n=\frac{-5±5}{-20}$$

Now solve the equation n=\frac{-5±5}{-20} when ± is plus. Add -5 to 5.

$$n=\frac{0}{-20}$$

Divide 0 by -20.

$$n=0$$

Now solve the equation n=\frac{-5±5}{-20} when ± is minus. Subtract 5 from -5.

$$n=-\frac{10}{-20}$$

Reduce the fraction \frac{-10}{-20} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

$$n=0 n=\frac{1}{2}$$

Steps for Completing the Square

Use the distributive property to multiply -5n by 2n-1.

$$-10n^{2}+5n=0$$

Divide both sides by -10.

$$\frac{-10n^{2}+5n}{-10}=\frac{0}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$n^{2}+\frac{5}{-10}n=\frac{0}{-10}$$

Reduce the fraction \frac{5}{-10} to lowest terms by extracting and canceling out 5.

$$n^{2}-\frac{1}{2}n=\frac{0}{-10}$$

Divide 0 by -10.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{1}{2}n+\left(-\frac{1}{4}\right)^{2}=\left(-\frac{1}{4}\right)^{2}$$

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{1}{16}$$

Factor n^{2}-\frac{1}{2}n+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{1}{4}\right)^{2}=\frac{1}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{4}\right)^{2}}=\sqrt{\frac{1}{16}}$$

Simplify.

$$n-\frac{1}{4}=\frac{1}{4} n-\frac{1}{4}=-\frac{1}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$n=\frac{1}{2} n=0$$