Solve left(n+1right)33=6n^2

$$\left( n+1 \right) 33 = 6 { n }^{ 2 }$$

Solve for n

$$n = \frac{\sqrt{209} + 11}{4} \approx 6.364208074$$

$$n=\frac{11-\sqrt{209}}{4}\approx -0.864208074$$

Steps Using the Quadratic Formula

Use the distributive property to multiply n+1 by 33.

$$33n+33=6n^{2}$$

Subtract 6n^{2} from both sides.

$$33n+33-6n^{2}=0$$

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.

$$-6n^{2}+33n+33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 33 for b, and 33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-33±\sqrt{33^{2}-4\left(-6\right)\times 33}}{2\left(-6\right)}$$

Square 33.

$$n=\frac{-33±\sqrt{1089-4\left(-6\right)\times 33}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$n=\frac{-33±\sqrt{1089+24\times 33}}{2\left(-6\right)}$$

Multiply 24 times 33.

$$n=\frac{-33±\sqrt{1089+792}}{2\left(-6\right)}$$

Add 1089 to 792.

$$n=\frac{-33±\sqrt{1881}}{2\left(-6\right)}$$

Take the square root of 1881.

$$n=\frac{-33±3\sqrt{209}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$n=\frac{-33±3\sqrt{209}}{-12}$$

Now solve the equation n=\frac{-33±3\sqrt{209}}{-12} when ± is plus. Add -33 to 3\sqrt{209}.

$$n=\frac{3\sqrt{209}-33}{-12}$$

Divide -33+3\sqrt{209} by -12.

$$n=\frac{11-\sqrt{209}}{4}$$

Now solve the equation n=\frac{-33±3\sqrt{209}}{-12} when ± is minus. Subtract 3\sqrt{209} from -33.

$$n=\frac{-3\sqrt{209}-33}{-12}$$

Divide -33-3\sqrt{209} by -12.

$$n=\frac{\sqrt{209}+11}{4}$$

The equation is now solved.

$$n=\frac{11-\sqrt{209}}{4} n=\frac{\sqrt{209}+11}{4}$$

Steps for Completing the Square

Use the distributive property to multiply n+1 by 33.

$$33n+33=6n^{2}$$

Subtract 6n^{2} from both sides.

$$33n+33-6n^{2}=0$$

Subtract 33 from both sides. Anything subtracted from zero gives its negation.

$$33n-6n^{2}=-33$$

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.

$$-6n^{2}+33n=-33$$

Divide both sides by -6.

$$\frac{-6n^{2}+33n}{-6}=-\frac{33}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$n^{2}+\frac{33}{-6}n=-\frac{33}{-6}$$

Reduce the fraction \frac{33}{-6} to lowest terms by extracting and canceling out 3.

$$n^{2}-\frac{11}{2}n=-\frac{33}{-6}$$

Reduce the fraction \frac{-33}{-6} to lowest terms by extracting and canceling out 3.

$$n^{2}-\frac{11}{2}n=\frac{11}{2}$$

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{11}{2}n+\left(-\frac{11}{4}\right)^{2}=\frac{11}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{11}{2}n+\frac{121}{16}=\frac{11}{2}+\frac{121}{16}$$

Add \frac{11}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{11}{2}n+\frac{121}{16}=\frac{209}{16}$$

Factor n^{2}-\frac{11}{2}n+\frac{121}{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}.

$$\left(n-\frac{11}{4}\right)^{2}=\frac{209}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{11}{4}\right)^{2}}=\sqrt{\frac{209}{16}}$$

Simplify.

$$n-\frac{11}{4}=\frac{\sqrt{209}}{4} n-\frac{11}{4}=-\frac{\sqrt{209}}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$n=\frac{\sqrt{209}+11}{4} n=\frac{11-\sqrt{209}}{4}$$