Solve 9x^2-1=(3x+1)(2x-3)

$$9 x ^ { 2 } - 1 = ( 3 x + 1 ) ( 2 x - 3 )$$

Solve for x

$$x=-2$$

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

Steps Using Factoring By Grouping

Use the distributive property to multiply 3x+1 by 2x-3 and combine like terms.

$$9x^{2}-1=6x^{2}-7x-3$$

Subtract 6x^{2} from both sides.

$$9x^{2}-1-6x^{2}=-7x-3$$

Combine 9x^{2} and -6x^{2} to get 3x^{2}.

$$3x^{2}-1=-7x-3$$

Add 7x to both sides.

$$3x^{2}-1+7x=-3$$

Add 3 to both sides.

$$3x^{2}-1+7x+3=0$$

Add -1 and 3 to get 2.

$$3x^{2}+2+7x=0$$

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

$$3x^{2}+7x+2=0$$

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

$$a+b=7 ab=3\times 2=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 7.

$$a=1 b=6$$

Rewrite 3x^{2}+7x+2 as \left(3x^{2}+x\right)+\left(6x+2\right).

$$\left(3x^{2}+x\right)+\left(6x+2\right)$$

Factor out x in the first and 2 in the second group.

$$x\left(3x+1\right)+2\left(3x+1\right)$$

Factor out common term 3x+1 by using distributive property.

$$\left(3x+1\right)\left(x+2\right)$$

To find equation solutions, solve 3x+1=0 and x+2=0.

$$x=-\frac{1}{3} x=-2$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x+1 by 2x-3 and combine like terms.

$$9x^{2}-1=6x^{2}-7x-3$$

Subtract 6x^{2} from both sides.

$$9x^{2}-1-6x^{2}=-7x-3$$

Combine 9x^{2} and -6x^{2} to get 3x^{2}.

$$3x^{2}-1=-7x-3$$

Add 7x to both sides.

$$3x^{2}-1+7x=-3$$

Add 3 to both sides.

$$3x^{2}-1+7x+3=0$$

Add -1 and 3 to get 2.

$$3x^{2}+2+7x=0$$

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.

$$3x^{2}+7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3\times 2}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-7±\sqrt{49-24}}{2\times 3}$$

Add 49 to -24.

$$x=\frac{-7±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$x=\frac{-7±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±5}{6}$$

Now solve the equation x=\frac{-7±5}{6} when ± is plus. Add -7 to 5.

$$x=-\frac{2}{6}$$

Reduce the fraction \frac{-2}{6} to lowest terms by extracting and canceling out 2.

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

Now solve the equation x=\frac{-7±5}{6} when ± is minus. Subtract 5 from -7.

$$x=-\frac{12}{6}$$

Divide -12 by 6.

$$x=-2$$

The equation is now solved.

$$x=-\frac{1}{3} x=-2$$

Steps for Completing the Square

Use the distributive property to multiply 3x+1 by 2x-3 and combine like terms.

$$9x^{2}-1=6x^{2}-7x-3$$

Subtract 6x^{2} from both sides.

$$9x^{2}-1-6x^{2}=-7x-3$$

Combine 9x^{2} and -6x^{2} to get 3x^{2}.

$$3x^{2}-1=-7x-3$$

Add 7x to both sides.

$$3x^{2}-1+7x=-3$$

Add 1 to both sides.

$$3x^{2}+7x=-3+1$$

Add -3 and 1 to get -2.

$$3x^{2}+7x=-2$$

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=-\frac{2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=-\frac{2}{3}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-\frac{2}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=-\frac{2}{3}+\frac{49}{36}$$

Add -\frac{2}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{25}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{5}{6} x+\frac{7}{6}=-\frac{5}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=-\frac{1}{3} x=-2$$