Solve (2)(3x-2)(4x+3)=(2x+1)(6x-5)+9

$$( 2 ) ( 3 x - 2 ) ( 4 x + 3 ) = ( 2 x + 1 ) ( 6 x - 5 ) + 9$$

Solve for x

$$x=\frac{\sqrt{201}}{12}-\frac{1}{4}\approx 0.931453907$$

$$x=-\frac{\sqrt{201}}{12}-\frac{1}{4}\approx -1.431453907$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2 by 3x-2.

$$\left(6x-4\right)\left(4x+3\right)=\left(2x+1\right)\left(6x-5\right)+9$$

Use the distributive property to multiply 6x-4 by 4x+3 and combine like terms.

$$24x^{2}+2x-12=\left(2x+1\right)\left(6x-5\right)+9$$

Use the distributive property to multiply 2x+1 by 6x-5 and combine like terms.

$$24x^{2}+2x-12=12x^{2}-4x-5+9$$

Add -5 and 9 to get 4.

$$24x^{2}+2x-12=12x^{2}-4x+4$$

Subtract 12x^{2} from both sides.

$$24x^{2}+2x-12-12x^{2}=-4x+4$$

Combine 24x^{2} and -12x^{2} to get 12x^{2}.

$$12x^{2}+2x-12=-4x+4$$

Add 4x to both sides.

$$12x^{2}+2x-12+4x=4$$

Combine 2x and 4x to get 6x.

$$12x^{2}+6x-12=4$$

Subtract 4 from both sides.

$$12x^{2}+6x-12-4=0$$

Subtract 4 from -12 to get -16.

$$12x^{2}+6x-16=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, 6 for b, and -16 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-6±\sqrt{6^{2}-4\times 12\left(-16\right)}}{2\times 12}$$

Square 6.

$$x=\frac{-6±\sqrt{36-4\times 12\left(-16\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-6±\sqrt{36-48\left(-16\right)}}{2\times 12}$$

Multiply -48 times -16.

$$x=\frac{-6±\sqrt{36+768}}{2\times 12}$$

Add 36 to 768.

$$x=\frac{-6±\sqrt{804}}{2\times 12}$$

Take the square root of 804.

$$x=\frac{-6±2\sqrt{201}}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{-6±2\sqrt{201}}{24}$$

Now solve the equation x=\frac{-6±2\sqrt{201}}{24} when ± is plus. Add -6 to 2\sqrt{201}.

$$x=\frac{2\sqrt{201}-6}{24}$$

Divide -6+2\sqrt{201} by 24.

$$x=\frac{\sqrt{201}}{12}-\frac{1}{4}$$

Now solve the equation x=\frac{-6±2\sqrt{201}}{24} when ± is minus. Subtract 2\sqrt{201} from -6.

$$x=\frac{-2\sqrt{201}-6}{24}$$

Divide -6-2\sqrt{201} by 24.

$$x=-\frac{\sqrt{201}}{12}-\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{201}}{12}-\frac{1}{4} x=-\frac{\sqrt{201}}{12}-\frac{1}{4}$$

Steps for Completing the Square

Use the distributive property to multiply 2 by 3x-2.

$$\left(6x-4\right)\left(4x+3\right)=\left(2x+1\right)\left(6x-5\right)+9$$

Use the distributive property to multiply 6x-4 by 4x+3 and combine like terms.

$$24x^{2}+2x-12=\left(2x+1\right)\left(6x-5\right)+9$$

Use the distributive property to multiply 2x+1 by 6x-5 and combine like terms.

$$24x^{2}+2x-12=12x^{2}-4x-5+9$$

Add -5 and 9 to get 4.

$$24x^{2}+2x-12=12x^{2}-4x+4$$

Subtract 12x^{2} from both sides.

$$24x^{2}+2x-12-12x^{2}=-4x+4$$

Combine 24x^{2} and -12x^{2} to get 12x^{2}.

$$12x^{2}+2x-12=-4x+4$$

Add 4x to both sides.

$$12x^{2}+2x-12+4x=4$$

Combine 2x and 4x to get 6x.

$$12x^{2}+6x-12=4$$

Add 12 to both sides.

$$12x^{2}+6x=4+12$$

Add 4 and 12 to get 16.

$$12x^{2}+6x=16$$

Divide both sides by 12.

$$\frac{12x^{2}+6x}{12}=\frac{16}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\frac{6}{12}x=\frac{16}{12}$$

Reduce the fraction \frac{6}{12} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{1}{2}x=\frac{16}{12}$$

Reduce the fraction \frac{16}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{1}{2}x=\frac{4}{3}$$

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=\frac{4}{3}+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{4}{3}+\frac{1}{16}$$

Add \frac{4}{3} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{67}{48}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{67}{48}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{67}{48}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{201}}{12} x+\frac{1}{4}=-\frac{\sqrt{201}}{12}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{201}}{12}-\frac{1}{4} x=-\frac{\sqrt{201}}{12}-\frac{1}{4}$$