Solve 21n^2-8=17n

$$21 n ^ { 2 } - 8 = 17 n$$

Solve for n

$$n=-\frac{1}{3}\approx -0.333333333$$

$$n = \frac{8}{7} = 1\frac{1}{7} \approx 1.142857143$$

Steps Using Factoring By Grouping

Subtract 17n from both sides.

$$21n^{2}-8-17n=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$21n^{2}-17n-8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 21n^{2}+an+bn-8. To find a and b, set up a system to be solved.

$$a+b=-17 ab=21\left(-8\right)=-168$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -168.

$$1,-168 2,-84 3,-56 4,-42 6,-28 7,-24 8,-21 12,-14$$

Calculate the sum for each pair.

$$1-168=-167 2-84=-82 3-56=-53 4-42=-38 6-28=-22 7-24=-17 8-21=-13 12-14=-2$$

The solution is the pair that gives sum -17.

$$a=-24 b=7$$

Rewrite 21n^{2}-17n-8 as \left(21n^{2}-24n\right)+\left(7n-8\right).

$$\left(21n^{2}-24n\right)+\left(7n-8\right)$$

Factor out 3n in 21n^{2}-24n.

$$3n\left(7n-8\right)+7n-8$$

Factor out common term 7n-8 by using distributive property.

$$\left(7n-8\right)\left(3n+1\right)$$

To find equation solutions, solve 7n-8=0 and 3n+1=0.

$$n=\frac{8}{7} n=-\frac{1}{3}$$

Steps Using the Quadratic Formula

Subtract 17n from both sides.

$$21n^{2}-8-17n=0$$

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.

$$21n^{2}-17n-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 21 for a, -17 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$n=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 21\left(-8\right)}}{2\times 21}$$

Square -17.

$$n=\frac{-\left(-17\right)±\sqrt{289-4\times 21\left(-8\right)}}{2\times 21}$$

Multiply -4 times 21.

$$n=\frac{-\left(-17\right)±\sqrt{289-84\left(-8\right)}}{2\times 21}$$

Multiply -84 times -8.

$$n=\frac{-\left(-17\right)±\sqrt{289+672}}{2\times 21}$$

Add 289 to 672.

$$n=\frac{-\left(-17\right)±\sqrt{961}}{2\times 21}$$

Take the square root of 961.

$$n=\frac{-\left(-17\right)±31}{2\times 21}$$

The opposite of -17 is 17.

$$n=\frac{17±31}{2\times 21}$$

Multiply 2 times 21.

$$n=\frac{17±31}{42}$$

Now solve the equation n=\frac{17±31}{42} when ± is plus. Add 17 to 31.

$$n=\frac{48}{42}$$

Reduce the fraction \frac{48}{42} to lowest terms by extracting and canceling out 6.

$$n=\frac{8}{7}$$

Now solve the equation n=\frac{17±31}{42} when ± is minus. Subtract 31 from 17.

$$n=-\frac{14}{42}$$

Reduce the fraction \frac{-14}{42} to lowest terms by extracting and canceling out 14.

$$n=-\frac{1}{3}$$

The equation is now solved.

$$n=\frac{8}{7} n=-\frac{1}{3}$$

Steps for Completing the Square

Subtract 17n from both sides.

$$21n^{2}-8-17n=0$$

Add 8 to both sides. Anything plus zero gives itself.

$$21n^{2}-17n=8$$

Divide both sides by 21.

$$\frac{21n^{2}-17n}{21}=\frac{8}{21}$$

Dividing by 21 undoes the multiplication by 21.

$$n^{2}-\frac{17}{21}n=\frac{8}{21}$$

Divide -\frac{17}{21}, the coefficient of the x term, by 2 to get -\frac{17}{42}. Then add the square of -\frac{17}{42} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$n^{2}-\frac{17}{21}n+\left(-\frac{17}{42}\right)^{2}=\frac{8}{21}+\left(-\frac{17}{42}\right)^{2}$$

Square -\frac{17}{42} by squaring both the numerator and the denominator of the fraction.

$$n^{2}-\frac{17}{21}n+\frac{289}{1764}=\frac{8}{21}+\frac{289}{1764}$$

Add \frac{8}{21} to \frac{289}{1764} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$n^{2}-\frac{17}{21}n+\frac{289}{1764}=\frac{961}{1764}$$

Factor n^{2}-\frac{17}{21}n+\frac{289}{1764}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(n-\frac{17}{42}\right)^{2}=\frac{961}{1764}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(n-\frac{17}{42}\right)^{2}}=\sqrt{\frac{961}{1764}}$$

Simplify.

$$n-\frac{17}{42}=\frac{31}{42} n-\frac{17}{42}=-\frac{31}{42}$$

Add \frac{17}{42} to both sides of the equation.

$$n=\frac{8}{7} n=-\frac{1}{3}$$