Solve x^2-9=(2x+6)(7x-2)

$$x ^ { 2 } - 9 = ( 2 x + 6 ) ( 7 x - 2 )$$

Solve for x

$$x=\frac{1}{13}\approx 0.076923077$$

$$x=-3$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x+6 by 7x-2 and combine like terms.

$$x^{2}-9=14x^{2}+38x-12$$

Subtract 14x^{2} from both sides.

$$x^{2}-9-14x^{2}=38x-12$$

Combine x^{2} and -14x^{2} to get -13x^{2}.

$$-13x^{2}-9=38x-12$$

Subtract 38x from both sides.

$$-13x^{2}-9-38x=-12$$

Add 12 to both sides.

$$-13x^{2}-9-38x+12=0$$

Add -9 and 12 to get 3.

$$-13x^{2}+3-38x=0$$

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

$$-13x^{2}-38x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -13x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=-38 ab=-13\times 3=-39$$

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 -39.

$$1,-39 3,-13$$

Calculate the sum for each pair.

$$1-39=-38 3-13=-10$$

The solution is the pair that gives sum -38.

$$a=1 b=-39$$

Rewrite -13x^{2}-38x+3 as \left(-13x^{2}+x\right)+\left(-39x+3\right).

$$\left(-13x^{2}+x\right)+\left(-39x+3\right)$$

Factor out -x in the first and -3 in the second group.

$$-x\left(13x-1\right)-3\left(13x-1\right)$$

Factor out common term 13x-1 by using distributive property.

$$\left(13x-1\right)\left(-x-3\right)$$

To find equation solutions, solve 13x-1=0 and -x-3=0.

$$x=\frac{1}{13} x=-3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x+6 by 7x-2 and combine like terms.

$$x^{2}-9=14x^{2}+38x-12$$

Subtract 14x^{2} from both sides.

$$x^{2}-9-14x^{2}=38x-12$$

Combine x^{2} and -14x^{2} to get -13x^{2}.

$$-13x^{2}-9=38x-12$$

Subtract 38x from both sides.

$$-13x^{2}-9-38x=-12$$

Add 12 to both sides.

$$-13x^{2}-9-38x+12=0$$

Add -9 and 12 to get 3.

$$-13x^{2}+3-38x=0$$

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.

$$-13x^{2}-38x+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -13 for a, -38 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-38\right)±\sqrt{\left(-38\right)^{2}-4\left(-13\right)\times 3}}{2\left(-13\right)}$$

Square -38.

$$x=\frac{-\left(-38\right)±\sqrt{1444-4\left(-13\right)\times 3}}{2\left(-13\right)}$$

Multiply -4 times -13.

$$x=\frac{-\left(-38\right)±\sqrt{1444+52\times 3}}{2\left(-13\right)}$$

Multiply 52 times 3.

$$x=\frac{-\left(-38\right)±\sqrt{1444+156}}{2\left(-13\right)}$$

Add 1444 to 156.

$$x=\frac{-\left(-38\right)±\sqrt{1600}}{2\left(-13\right)}$$

Take the square root of 1600.

$$x=\frac{-\left(-38\right)±40}{2\left(-13\right)}$$

The opposite of -38 is 38.

$$x=\frac{38±40}{2\left(-13\right)}$$

Multiply 2 times -13.

$$x=\frac{38±40}{-26}$$

Now solve the equation x=\frac{38±40}{-26} when ± is plus. Add 38 to 40.

$$x=\frac{78}{-26}$$

Divide 78 by -26.

$$x=-3$$

Now solve the equation x=\frac{38±40}{-26} when ± is minus. Subtract 40 from 38.

$$x=-\frac{2}{-26}$$

Reduce the fraction \frac{-2}{-26} to lowest terms by extracting and canceling out 2.

$$x=\frac{1}{13}$$

The equation is now solved.

$$x=-3 x=\frac{1}{13}$$

Steps for Completing the Square

Use the distributive property to multiply 2x+6 by 7x-2 and combine like terms.

$$x^{2}-9=14x^{2}+38x-12$$

Subtract 14x^{2} from both sides.

$$x^{2}-9-14x^{2}=38x-12$$

Combine x^{2} and -14x^{2} to get -13x^{2}.

$$-13x^{2}-9=38x-12$$

Subtract 38x from both sides.

$$-13x^{2}-9-38x=-12$$

Add 9 to both sides.

$$-13x^{2}-38x=-12+9$$

Add -12 and 9 to get -3.

$$-13x^{2}-38x=-3$$

Divide both sides by -13.

$$\frac{-13x^{2}-38x}{-13}=-\frac{3}{-13}$$

Dividing by -13 undoes the multiplication by -13.

$$x^{2}+\left(-\frac{38}{-13}\right)x=-\frac{3}{-13}$$

Divide -38 by -13.

$$x^{2}+\frac{38}{13}x=-\frac{3}{-13}$$

Divide -3 by -13.

$$x^{2}+\frac{38}{13}x=\frac{3}{13}$$

Divide \frac{38}{13}, the coefficient of the x term, by 2 to get \frac{19}{13}. Then add the square of \frac{19}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{38}{13}x+\left(\frac{19}{13}\right)^{2}=\frac{3}{13}+\left(\frac{19}{13}\right)^{2}$$

Square \frac{19}{13} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{38}{13}x+\frac{361}{169}=\frac{3}{13}+\frac{361}{169}$$

Add \frac{3}{13} to \frac{361}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{38}{13}x+\frac{361}{169}=\frac{400}{169}$$

Factor x^{2}+\frac{38}{13}x+\frac{361}{169}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{13}\right)^{2}=\frac{400}{169}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{13}\right)^{2}}=\sqrt{\frac{400}{169}}$$

Simplify.

$$x+\frac{19}{13}=\frac{20}{13} x+\frac{19}{13}=-\frac{20}{13}$$

Subtract \frac{19}{13} from both sides of the equation.

$$x=\frac{1}{13} x=-3$$