Solve 5x^2-x-2184=0

$$5 { x }^{ 2 } -x-2184=0$$

Solve for x

$$x = -\frac{104}{5} = -20\frac{4}{5} = -20.8$$

$$x=21$$

Steps Using Factoring By Grouping

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-2184. To find a and b, set up a system to be solved.

$$a+b=-1 ab=5\left(-2184\right)=-10920$$

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 -10920.

$$1,-10920 2,-5460 3,-3640 4,-2730 5,-2184 6,-1820 7,-1560 8,-1365 10,-1092 12,-910 13,-840 14,-780 15,-728 20,-546 21,-520 24,-455 26,-420 28,-390 30,-364 35,-312 39,-280 40,-273 42,-260 52,-210 56,-195 60,-182 65,-168 70,-156 78,-140 84,-130 91,-120 104,-105$$

Calculate the sum for each pair.

$$1-10920=-10919 2-5460=-5458 3-3640=-3637 4-2730=-2726 5-2184=-2179 6-1820=-1814 7-1560=-1553 8-1365=-1357 10-1092=-1082 12-910=-898 13-840=-827 14-780=-766 15-728=-713 20-546=-526 21-520=-499 24-455=-431 26-420=-394 28-390=-362 30-364=-334 35-312=-277 39-280=-241 40-273=-233 42-260=-218 52-210=-158 56-195=-139 60-182=-122 65-168=-103 70-156=-86 78-140=-62 84-130=-46 91-120=-29 104-105=-1$$

The solution is the pair that gives sum -1.

$$a=-105 b=104$$

Rewrite 5x^{2}-x-2184 as \left(5x^{2}-105x\right)+\left(104x-2184\right).

$$\left(5x^{2}-105x\right)+\left(104x-2184\right)$$

Factor out 5x in the first and 104 in the second group.

$$5x\left(x-21\right)+104\left(x-21\right)$$

Factor out common term x-21 by using distributive property.

$$\left(x-21\right)\left(5x+104\right)$$

To find equation solutions, solve x-21=0 and 5x+104=0.

$$x=21 x=-\frac{104}{5}$$

Steps Using the Quadratic Formula

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.

$$5x^{2}-x-2184=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -1 for b, and -2184 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 5\left(-2184\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-1\right)±\sqrt{1-20\left(-2184\right)}}{2\times 5}$$

Multiply -20 times -2184.

$$x=\frac{-\left(-1\right)±\sqrt{1+43680}}{2\times 5}$$

Add 1 to 43680.

$$x=\frac{-\left(-1\right)±\sqrt{43681}}{2\times 5}$$

Take the square root of 43681.

$$x=\frac{-\left(-1\right)±209}{2\times 5}$$

The opposite of -1 is 1.

$$x=\frac{1±209}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{1±209}{10}$$

Now solve the equation x=\frac{1±209}{10} when ± is plus. Add 1 to 209.

$$x=\frac{210}{10}$$

Divide 210 by 10.

$$x=21$$

Now solve the equation x=\frac{1±209}{10} when ± is minus. Subtract 209 from 1.

$$x=-\frac{208}{10}$$

Reduce the fraction \frac{-208}{10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{104}{5}$$

The equation is now solved.

$$x=21 x=-\frac{104}{5}$$

Steps for Completing the Square

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.

$$5x^{2}-x-2184=0$$

Add 2184 to both sides of the equation.

$$5x^{2}-x-2184-\left(-2184\right)=-\left(-2184\right)$$

Subtracting -2184 from itself leaves 0.

$$5x^{2}-x=-\left(-2184\right)$$

Subtract -2184 from 0.

$$5x^{2}-x=2184$$

Divide both sides by 5.

$$\frac{5x^{2}-x}{5}=\frac{2184}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{1}{5}x=\frac{2184}{5}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{2184}{5}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{2184}{5}+\frac{1}{100}$$

Add \frac{2184}{5} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{43681}{100}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{43681}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{43681}{100}}$$

Simplify.

$$x-\frac{1}{10}=\frac{209}{10} x-\frac{1}{10}=-\frac{209}{10}$$

Add \frac{1}{10} to both sides of the equation.

$$x=21 x=-\frac{104}{5}$$