Solve -8n(4n-7)=0

$$- 8 n ( 4 n - 7 ) = 0$$

Solve for n

$$n = \frac{7}{4} = 1\frac{3}{4} = 1.75$$

$$n=0$$

Steps Using Factoring

Use the distributive property to multiply -8n by 4n-7.

$$-32n^{2}+56n=0$$

Factor out n.

$$n\left(-32n+56\right)=0$$

To find equation solutions, solve n=0 and -32n+56=0.

$$n=0 n=\frac{7}{4}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -8n by 4n-7.

$$-32n^{2}+56n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -32 for a, 56 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-56±\sqrt{56^{2}}}{2\left(-32\right)}$$

Take the square root of 56^{2}.

$$n=\frac{-56±56}{2\left(-32\right)}$$

Multiply 2 times -32.

$$n=\frac{-56±56}{-64}$$

Now solve the equation n=\frac{-56±56}{-64} when ± is plus. Add -56 to 56.

$$n=\frac{0}{-64}$$

Divide 0 by -64.

$$n=0$$

Now solve the equation n=\frac{-56±56}{-64} when ± is minus. Subtract 56 from -56.

$$n=-\frac{112}{-64}$$

Reduce the fraction \frac{-112}{-64} to lowest terms by extracting and canceling out 16.

$$n=\frac{7}{4}$$

The equation is now solved.

$$n=0 n=\frac{7}{4}$$

Steps for Completing the Square

Use the distributive property to multiply -8n by 4n-7.

$$-32n^{2}+56n=0$$

Divide both sides by -32.

$$\frac{-32n^{2}+56n}{-32}=\frac{0}{-32}$$

Dividing by -32 undoes the multiplication by -32.

$$n^{2}+\frac{56}{-32}n=\frac{0}{-32}$$

Reduce the fraction \frac{56}{-32} to lowest terms by extracting and canceling out 8.

$$n^{2}-\frac{7}{4}n=\frac{0}{-32}$$

Divide 0 by -32.

$$n^{2}-\frac{7}{4}n=0$$

Divide -\frac{7}{4}, the coefficient of the x term, by 2 to get -\frac{7}{8}. Then add the square of -\frac{7}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{7}{4}n+\left(-\frac{7}{8}\right)^{2}=\left(-\frac{7}{8}\right)^{2}$$

Square -\frac{7}{8} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{7}{4}n+\frac{49}{64}=\frac{49}{64}$$

Factor n^{2}-\frac{7}{4}n+\frac{49}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{7}{8}\right)^{2}=\frac{49}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{7}{8}\right)^{2}}=\sqrt{\frac{49}{64}}$$

Simplify.

$$n-\frac{7}{8}=\frac{7}{8} n-\frac{7}{8}=-\frac{7}{8}$$

Add \frac{7}{8} to both sides of the equation.

$$n=\frac{7}{4} n=0$$