Solve 0=-1/5left(3.2-nright)^2+16

$$0 = - \frac{ 1 }{ 5 } { \left(3.2-n \right) }^{ 2 } +16$$

Solve for n

$$n=3.2-4\sqrt{5}\approx -5.74427191$$

$$n=4\sqrt{5}+3.2\approx 12.14427191$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3.2-n\right)^{2}.

$$0=-\frac{1}{5}\left(10.24-6.4n+n^{2}\right)+16$$

Use the distributive property to multiply -\frac{1}{5} by 10.24-6.4n+n^{2}.

$$0=-\frac{256}{125}+\frac{32}{25}n-\frac{1}{5}n^{2}+16$$

Add -\frac{256}{125} and 16 to get \frac{1744}{125}.

$$0=\frac{1744}{125}+\frac{32}{25}n-\frac{1}{5}n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1744}{125}+\frac{32}{25}n-\frac{1}{5}n^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-\frac{1}{5}n^{2}+\frac{32}{25}n+\frac{1744}{125}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{5} for a, \frac{32}{25} for b, and \frac{1744}{125} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\frac{32}{25}±\sqrt{\left(\frac{32}{25}\right)^{2}-4\left(-\frac{1}{5}\right)\times \frac{1744}{125}}}{2\left(-\frac{1}{5}\right)}$$

Square \frac{32}{25} by squaring both the numerator and the denominator of the fraction.

$$n=\frac{-\frac{32}{25}±\sqrt{\frac{1024}{625}-4\left(-\frac{1}{5}\right)\times \frac{1744}{125}}}{2\left(-\frac{1}{5}\right)}$$

Multiply -4 times -\frac{1}{5}.

$$n=\frac{-\frac{32}{25}±\sqrt{\frac{1024}{625}+\frac{4}{5}\times \frac{1744}{125}}}{2\left(-\frac{1}{5}\right)}$$

Multiply \frac{4}{5} times \frac{1744}{125} by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.

$$n=\frac{-\frac{32}{25}±\sqrt{\frac{1024+6976}{625}}}{2\left(-\frac{1}{5}\right)}$$

Add \frac{1024}{625} to \frac{6976}{625} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n=\frac{-\frac{32}{25}±\sqrt{\frac{64}{5}}}{2\left(-\frac{1}{5}\right)}$$

Take the square root of \frac{64}{5}.

$$n=\frac{-\frac{32}{25}±\frac{8\sqrt{5}}{5}}{2\left(-\frac{1}{5}\right)}$$

Multiply 2 times -\frac{1}{5}.

$$n=\frac{-\frac{32}{25}±\frac{8\sqrt{5}}{5}}{-\frac{2}{5}}$$

Now solve the equation n=\frac{-\frac{32}{25}±\frac{8\sqrt{5}}{5}}{-\frac{2}{5}} when ± is plus. Add -\frac{32}{25} to \frac{8\sqrt{5}}{5}.

$$n=\frac{\frac{8\sqrt{5}}{5}-\frac{32}{25}}{-\frac{2}{5}}$$

Divide -\frac{32}{25}+\frac{8\sqrt{5}}{5} by -\frac{2}{5} by multiplying -\frac{32}{25}+\frac{8\sqrt{5}}{5} by the reciprocal of -\frac{2}{5}.

$$n=\frac{16}{5}-4\sqrt{5}$$

Now solve the equation n=\frac{-\frac{32}{25}±\frac{8\sqrt{5}}{5}}{-\frac{2}{5}} when ± is minus. Subtract \frac{8\sqrt{5}}{5} from -\frac{32}{25}.

$$n=\frac{-\frac{8\sqrt{5}}{5}-\frac{32}{25}}{-\frac{2}{5}}$$

Divide -\frac{32}{25}-\frac{8\sqrt{5}}{5} by -\frac{2}{5} by multiplying -\frac{32}{25}-\frac{8\sqrt{5}}{5} by the reciprocal of -\frac{2}{5}.

$$n=4\sqrt{5}+\frac{16}{5}$$

The equation is now solved.

$$n=\frac{16}{5}-4\sqrt{5} n=4\sqrt{5}+\frac{16}{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3.2-n\right)^{2}.

$$0=-\frac{1}{5}\left(10.24-6.4n+n^{2}\right)+16$$

Use the distributive property to multiply -\frac{1}{5} by 10.24-6.4n+n^{2}.

$$0=-\frac{256}{125}+\frac{32}{25}n-\frac{1}{5}n^{2}+16$$

Add -\frac{256}{125} and 16 to get \frac{1744}{125}.

$$0=\frac{1744}{125}+\frac{32}{25}n-\frac{1}{5}n^{2}$$

Swap sides so that all variable terms are on the left hand side.

$$\frac{1744}{125}+\frac{32}{25}n-\frac{1}{5}n^{2}=0$$

Subtract \frac{1744}{125} from both sides. Anything subtracted from zero gives its negation.

$$\frac{32}{25}n-\frac{1}{5}n^{2}=-\frac{1744}{125}$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-\frac{1}{5}n^{2}+\frac{32}{25}n=-\frac{1744}{125}$$

Multiply both sides by -5.

$$\frac{-\frac{1}{5}n^{2}+\frac{32}{25}n}{-\frac{1}{5}}=-\frac{\frac{1744}{125}}{-\frac{1}{5}}$$

Dividing by -\frac{1}{5} undoes the multiplication by -\frac{1}{5}.

$$n^{2}+\frac{\frac{32}{25}}{-\frac{1}{5}}n=-\frac{\frac{1744}{125}}{-\frac{1}{5}}$$

Divide \frac{32}{25} by -\frac{1}{5} by multiplying \frac{32}{25} by the reciprocal of -\frac{1}{5}.

$$n^{2}-\frac{32}{5}n=-\frac{\frac{1744}{125}}{-\frac{1}{5}}$$

Divide -\frac{1744}{125} by -\frac{1}{5} by multiplying -\frac{1744}{125} by the reciprocal of -\frac{1}{5}.

$$n^{2}-\frac{32}{5}n=\frac{1744}{25}$$

Divide -\frac{32}{5}, the coefficient of the x term, by 2 to get -\frac{16}{5}. Then add the square of -\frac{16}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{32}{5}n+\left(-\frac{16}{5}\right)^{2}=\frac{1744}{25}+\left(-\frac{16}{5}\right)^{2}$$

Square -\frac{16}{5} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{32}{5}n+\frac{256}{25}=\frac{1744+256}{25}$$

Add \frac{1744}{25} to \frac{256}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{32}{5}n+\frac{256}{25}=80$$

Factor n^{2}-\frac{32}{5}n+\frac{256}{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}.

$$\left(n-\frac{16}{5}\right)^{2}=80$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{16}{5}\right)^{2}}=\sqrt{80}$$

Simplify.

$$n-\frac{16}{5}=4\sqrt{5} n-\frac{16}{5}=-4\sqrt{5}$$

Add \frac{16}{5} to both sides of the equation.

$$n=4\sqrt{5}+\frac{16}{5} n=\frac{16}{5}-4\sqrt{5}$$