Solve -7n(5n-9)=0

$$- 7 n ( 5 n - 9 ) = 0$$

Solve for n

$$n = \frac{9}{5} = 1\frac{4}{5} = 1.8$$

$$n=0$$

Steps Using Factoring

Use the distributive property to multiply -7n by 5n-9.

$$-35n^{2}+63n=0$$

Factor out n.

$$n\left(-35n+63\right)=0$$

To find equation solutions, solve n=0 and -35n+63=0.

$$n=0 n=\frac{9}{5}$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -7n by 5n-9.

$$-35n^{2}+63n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -35 for a, 63 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-63±\sqrt{63^{2}}}{2\left(-35\right)}$$

Take the square root of 63^{2}.

$$n=\frac{-63±63}{2\left(-35\right)}$$

Multiply 2 times -35.

$$n=\frac{-63±63}{-70}$$

Now solve the equation n=\frac{-63±63}{-70} when ± is plus. Add -63 to 63.

$$n=\frac{0}{-70}$$

Divide 0 by -70.

$$n=0$$

Now solve the equation n=\frac{-63±63}{-70} when ± is minus. Subtract 63 from -63.

$$n=-\frac{126}{-70}$$

Reduce the fraction \frac{-126}{-70} to lowest terms by extracting and canceling out 14.

$$n=\frac{9}{5}$$

The equation is now solved.

$$n=0 n=\frac{9}{5}$$

Steps for Completing the Square

Use the distributive property to multiply -7n by 5n-9.

$$-35n^{2}+63n=0$$

Divide both sides by -35.

$$\frac{-35n^{2}+63n}{-35}=\frac{0}{-35}$$

Dividing by -35 undoes the multiplication by -35.

$$n^{2}+\frac{63}{-35}n=\frac{0}{-35}$$

Reduce the fraction \frac{63}{-35} to lowest terms by extracting and canceling out 7.

$$n^{2}-\frac{9}{5}n=\frac{0}{-35}$$

Divide 0 by -35.

$$n^{2}-\frac{9}{5}n=0$$

Divide -\frac{9}{5}, the coefficient of the x term, by 2 to get -\frac{9}{10}. Then add the square of -\frac{9}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{9}{5}n+\left(-\frac{9}{10}\right)^{2}=\left(-\frac{9}{10}\right)^{2}$$

Square -\frac{9}{10} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{9}{5}n+\frac{81}{100}=\frac{81}{100}$$

Factor n^{2}-\frac{9}{5}n+\frac{81}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{9}{10}\right)^{2}=\frac{81}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{9}{10}\right)^{2}}=\sqrt{\frac{81}{100}}$$

Simplify.

$$n-\frac{9}{10}=\frac{9}{10} n-\frac{9}{10}=-\frac{9}{10}$$

Add \frac{9}{10} to both sides of the equation.

$$n=\frac{9}{5} n=0$$