Solve 1/2*22/7*n^2=1/4*22/7*(n+2)

$$\frac { 1 } { 2 } \times \frac { 22 } { 7 } \times n ^ { 2 } = \frac { 1 } { 4 } \times \frac { 22 } { 7 } \times ( n + 2 )$$

Solve for n

$$n = \frac{\sqrt{17} + 1}{4} \approx 1.280776406$$

$$n=\frac{1-\sqrt{17}}{4}\approx -0.780776406$$

Steps Using the Quadratic Formula

Cancel out \frac{22}{7} on both sides.

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

Use the distributive property to multiply \frac{1}{4} by n+2.

$$\frac{1}{2}n^{2}=\frac{1}{4}n+\frac{1}{2}$$

Subtract \frac{1}{4}n from both sides.

$$\frac{1}{2}n^{2}-\frac{1}{4}n=\frac{1}{2}$$

Subtract \frac{1}{2} from both sides.

$$\frac{1}{2}n^{2}-\frac{1}{4}n-\frac{1}{2}=0$$

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

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

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

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

Multiply -4 times \frac{1}{2}.

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

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

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

Add \frac{1}{16} to 1.

$$n=\frac{-\left(-\frac{1}{4}\right)±\sqrt{\frac{17}{16}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{17}{16}.

$$n=\frac{-\left(-\frac{1}{4}\right)±\frac{\sqrt{17}}{4}}{2\times \frac{1}{2}}$$

The opposite of -\frac{1}{4} is \frac{1}{4}.

$$n=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{2\times \frac{1}{2}}$$

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

$$n=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{1}$$

Now solve the equation n=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{1} when ± is plus. Add \frac{1}{4} to \frac{\sqrt{17}}{4}.

$$n=\frac{\sqrt{17}+1}{4}$$

Now solve the equation n=\frac{\frac{1}{4}±\frac{\sqrt{17}}{4}}{1} when ± is minus. Subtract \frac{\sqrt{17}}{4} from \frac{1}{4}.

$$n=\frac{1-\sqrt{17}}{4}$$

The equation is now solved.

$$n=\frac{\sqrt{17}+1}{4} n=\frac{1-\sqrt{17}}{4}$$

Steps for Completing the Square

Cancel out \frac{22}{7} on both sides.

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

Use the distributive property to multiply \frac{1}{4} by n+2.

$$\frac{1}{2}n^{2}=\frac{1}{4}n+\frac{1}{2}$$

Subtract \frac{1}{4}n from both sides.

$$\frac{1}{2}n^{2}-\frac{1}{4}n=\frac{1}{2}$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}n^{2}-\frac{1}{4}n}{\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

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

Divide -\frac{1}{4} by \frac{1}{2} by multiplying -\frac{1}{4} by the reciprocal of \frac{1}{2}.

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

Divide \frac{1}{2} by \frac{1}{2} by multiplying \frac{1}{2} by the reciprocal of \frac{1}{2}.

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

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.

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

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

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

Add 1 to \frac{1}{16}.

$$n^{2}-\frac{1}{2}n+\frac{1}{16}=\frac{17}{16}$$

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}.

$$\left(n-\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$n-\frac{1}{4}=\frac{\sqrt{17}}{4} n-\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

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

$$n=\frac{\sqrt{17}+1}{4} n=\frac{1-\sqrt{17}}{4}$$