Solve 9x^2-1=(2x-5)(3x+1)

$$9 x ^ { 2 } - 1 = ( 2 x - 5 ) ( 3 x + 1 )$$

Solve for x

$$x=-4$$

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

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x-5 by 3x+1 and combine like terms.

$$9x^{2}-1=6x^{2}-13x-5$$

Subtract 6x^{2} from both sides.

$$9x^{2}-1-6x^{2}=-13x-5$$

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

$$3x^{2}-1=-13x-5$$

Add 13x to both sides.

$$3x^{2}-1+13x=-5$$

Add 5 to both sides.

$$3x^{2}-1+13x+5=0$$

Add -1 and 5 to get 4.

$$3x^{2}+4+13x=0$$

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

$$3x^{2}+13x+4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+4. To find a and b, set up a system to be solved.

$$a+b=13 ab=3\times 4=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=1 b=12$$

Rewrite 3x^{2}+13x+4 as \left(3x^{2}+x\right)+\left(12x+4\right).

$$\left(3x^{2}+x\right)+\left(12x+4\right)$$

Factor out x in the first and 4 in the second group.

$$x\left(3x+1\right)+4\left(3x+1\right)$$

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

$$\left(3x+1\right)\left(x+4\right)$$

To find equation solutions, solve 3x+1=0 and x+4=0.

$$x=-\frac{1}{3} x=-4$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x-5 by 3x+1 and combine like terms.

$$9x^{2}-1=6x^{2}-13x-5$$

Subtract 6x^{2} from both sides.

$$9x^{2}-1-6x^{2}=-13x-5$$

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

$$3x^{2}-1=-13x-5$$

Add 13x to both sides.

$$3x^{2}-1+13x=-5$$

Add 5 to both sides.

$$3x^{2}-1+13x+5=0$$

Add -1 and 5 to get 4.

$$3x^{2}+4+13x=0$$

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.

$$3x^{2}+13x+4=0$$

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

$$x=\frac{-13±\sqrt{13^{2}-4\times 3\times 4}}{2\times 3}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 3\times 4}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-13±\sqrt{169-12\times 4}}{2\times 3}$$

Multiply -12 times 4.

$$x=\frac{-13±\sqrt{169-48}}{2\times 3}$$

Add 169 to -48.

$$x=\frac{-13±\sqrt{121}}{2\times 3}$$

Take the square root of 121.

$$x=\frac{-13±11}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-13±11}{6}$$

Now solve the equation x=\frac{-13±11}{6} when ± is plus. Add -13 to 11.

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

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{-13±11}{6} when ± is minus. Subtract 11 from -13.

$$x=-\frac{24}{6}$$

Divide -24 by 6.

$$x=-4$$

The equation is now solved.

$$x=-\frac{1}{3} x=-4$$

Steps for Completing the Square

Use the distributive property to multiply 2x-5 by 3x+1 and combine like terms.

$$9x^{2}-1=6x^{2}-13x-5$$

Subtract 6x^{2} from both sides.

$$9x^{2}-1-6x^{2}=-13x-5$$

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

$$3x^{2}-1=-13x-5$$

Add 13x to both sides.

$$3x^{2}-1+13x=-5$$

Add 1 to both sides.

$$3x^{2}+13x=-5+1$$

Add -5 and 1 to get -4.

$$3x^{2}+13x=-4$$

Divide both sides by 3.

$$\frac{3x^{2}+13x}{3}=-\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{13}{3}x=-\frac{4}{3}$$

Divide \frac{13}{3}, the coefficient of the x term, by 2 to get \frac{13}{6}. Then add the square of \frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{3}x+\left(\frac{13}{6}\right)^{2}=-\frac{4}{3}+\left(\frac{13}{6}\right)^{2}$$

Square \frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=-\frac{4}{3}+\frac{169}{36}$$

Add -\frac{4}{3} to \frac{169}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{13}{3}x+\frac{169}{36}=\frac{121}{36}$$

Factor x^{2}+\frac{13}{3}x+\frac{169}{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}.

$$\left(x+\frac{13}{6}\right)^{2}=\frac{121}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{6}\right)^{2}}=\sqrt{\frac{121}{36}}$$

Simplify.

$$x+\frac{13}{6}=\frac{11}{6} x+\frac{13}{6}=-\frac{11}{6}$$

Subtract \frac{13}{6} from both sides of the equation.

$$x=-\frac{1}{3} x=-4$$