Solve 5n^2-30n^2+40n=0

$$5 n ^ { 2 } - 30 n ^ { 2 } + 40 n = 0$$

Solve for n

$$n = \frac{8}{5} = 1\frac{3}{5} = 1.6$$

$$n=0$$

Steps Using Factoring

Factor out n.

$$n\left(5n-30n+40\right)=0$$

To find equation solutions, solve n=0 and 5n-30n+40=0.

$$n=0 n=\frac{8}{5}$$

Steps Using the Quadratic Formula

Combine 5n^{2} and -30n^{2} to get -25n^{2}.

$$-25n^{2}+40n=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -25 for a, 40 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-40±\sqrt{40^{2}}}{2\left(-25\right)}$$

Take the square root of 40^{2}.

$$n=\frac{-40±40}{2\left(-25\right)}$$

Multiply 2 times -25.

$$n=\frac{-40±40}{-50}$$

Now solve the equation n=\frac{-40±40}{-50} when ± is plus. Add -40 to 40.

$$n=\frac{0}{-50}$$

Divide 0 by -50.

$$n=0$$

Now solve the equation n=\frac{-40±40}{-50} when ± is minus. Subtract 40 from -40.

$$n=-\frac{80}{-50}$$

Reduce the fraction \frac{-80}{-50} to lowest terms by extracting and canceling out 10.

$$n=\frac{8}{5}$$

The equation is now solved.

$$n=0 n=\frac{8}{5}$$

Steps for Completing the Square

Combine 5n^{2} and -30n^{2} to get -25n^{2}.

$$-25n^{2}+40n=0$$

Divide both sides by -25.

$$\frac{-25n^{2}+40n}{-25}=\frac{0}{-25}$$

Dividing by -25 undoes the multiplication by -25.

$$n^{2}+\frac{40}{-25}n=\frac{0}{-25}$$

Reduce the fraction \frac{40}{-25} to lowest terms by extracting and canceling out 5.

$$n^{2}-\frac{8}{5}n=\frac{0}{-25}$$

Divide 0 by -25.

$$n^{2}-\frac{8}{5}n=0$$

Divide -\frac{8}{5}, the coefficient of the x term, by 2 to get -\frac{4}{5}. Then add the square of -\frac{4}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{8}{5}n+\left(-\frac{4}{5}\right)^{2}=\left(-\frac{4}{5}\right)^{2}$$

Square -\frac{4}{5} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{8}{5}n+\frac{16}{25}=\frac{16}{25}$$

Factor n^{2}-\frac{8}{5}n+\frac{16}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{4}{5}\right)^{2}=\frac{16}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{4}{5}\right)^{2}}=\sqrt{\frac{16}{25}}$$

Simplify.

$$n-\frac{4}{5}=\frac{4}{5} n-\frac{4}{5}=-\frac{4}{5}$$

Add \frac{4}{5} to both sides of the equation.

$$n=\frac{8}{5} n=0$$