Solve (2*6+2)!/(2*6)!=4x^2+5x+2

$$\frac{ (2 \times 6+2) ! }{ (2 \times 6) ! } =4 { x }^{ 2 } +5x+2$$

Solve for x

$$x = \frac{\sqrt{2905} - 5}{8} \approx 6.112256489$$

$$x=\frac{-\sqrt{2905}-5}{8}\approx -7.362256489$$

Steps Using the Quadratic Formula

Multiply 2 and 6 to get 12.

$$\frac{\left(12+2\right)!}{\left(2\times 6\right)!}=4x^{2}+5x+2$$

Add 12 and 2 to get 14.

$$\frac{14!}{\left(2\times 6\right)!}=4x^{2}+5x+2$$

The factorial of 14 is 87178291200.

$$\frac{87178291200}{\left(2\times 6\right)!}=4x^{2}+5x+2$$

Multiply 2 and 6 to get 12.

$$\frac{87178291200}{12!}=4x^{2}+5x+2$$

The factorial of 12 is 479001600.

$$\frac{87178291200}{479001600}=4x^{2}+5x+2$$

Divide 87178291200 by 479001600 to get 182.

$$182=4x^{2}+5x+2$$

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

$$4x^{2}+5x+2=182$$

Subtract 182 from both sides.

$$4x^{2}+5x+2-182=0$$

Subtract 182 from 2 to get -180.

$$4x^{2}+5x-180=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 5 for b, and -180 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\times 4\left(-180\right)}}{2\times 4}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\times 4\left(-180\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-5±\sqrt{25-16\left(-180\right)}}{2\times 4}$$

Multiply -16 times -180.

$$x=\frac{-5±\sqrt{25+2880}}{2\times 4}$$

Add 25 to 2880.

$$x=\frac{-5±\sqrt{2905}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-5±\sqrt{2905}}{8}$$

Now solve the equation x=\frac{-5±\sqrt{2905}}{8} when ± is plus. Add -5 to \sqrt{2905}.

$$x=\frac{\sqrt{2905}-5}{8}$$

Now solve the equation x=\frac{-5±\sqrt{2905}}{8} when ± is minus. Subtract \sqrt{2905} from -5.

$$x=\frac{-\sqrt{2905}-5}{8}$$

The equation is now solved.

$$x=\frac{\sqrt{2905}-5}{8} x=\frac{-\sqrt{2905}-5}{8}$$

Steps for Completing the Square

Multiply 2 and 6 to get 12.

$$\frac{\left(12+2\right)!}{\left(2\times 6\right)!}=4x^{2}+5x+2$$

Add 12 and 2 to get 14.

$$\frac{14!}{\left(2\times 6\right)!}=4x^{2}+5x+2$$

The factorial of 14 is 87178291200.

$$\frac{87178291200}{\left(2\times 6\right)!}=4x^{2}+5x+2$$

Multiply 2 and 6 to get 12.

$$\frac{87178291200}{12!}=4x^{2}+5x+2$$

The factorial of 12 is 479001600.

$$\frac{87178291200}{479001600}=4x^{2}+5x+2$$

Divide 87178291200 by 479001600 to get 182.

$$182=4x^{2}+5x+2$$

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

$$4x^{2}+5x+2=182$$

Subtract 2 from both sides.

$$4x^{2}+5x=182-2$$

Subtract 2 from 182 to get 180.

$$4x^{2}+5x=180$$

Divide both sides by 4.

$$\frac{4x^{2}+5x}{4}=\frac{180}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\frac{5}{4}x=\frac{180}{4}$$

Divide 180 by 4.

$$x^{2}+\frac{5}{4}x=45$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=45+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=45+\frac{25}{64}$$

Add 45 to \frac{25}{64}.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{2905}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{2905}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{2905}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{\sqrt{2905}}{8} x+\frac{5}{8}=-\frac{\sqrt{2905}}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

$$x=\frac{\sqrt{2905}-5}{8} x=\frac{-\sqrt{2905}-5}{8}$$