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

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

Solve for x

$$x=-8$$

$$x=1$$

Steps Using the Quadratic Formula

Consider \left(3x-1\right)\left(3x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(3x\right)^{2}-1-\left(5x-4\right)\left(2x+3\right)=3$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-1-\left(5x-4\right)\left(2x+3\right)=3$$

Calculate 3 to the power of 2 and get 9.

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

Use the distributive property to multiply 5x-4 by 2x+3 and combine like terms.

$$9x^{2}-1-\left(10x^{2}+7x-12\right)=3$$

To find the opposite of 10x^{2}+7x-12, find the opposite of each term.

$$9x^{2}-1-10x^{2}-7x+12=3$$

Combine 9x^{2} and -10x^{2} to get -x^{2}.

$$-x^{2}-1-7x+12=3$$

Add -1 and 12 to get 11.

$$-x^{2}+11-7x=3$$

Subtract 3 from both sides.

$$-x^{2}+11-7x-3=0$$

Subtract 3 from 11 to get 8.

$$-x^{2}+8-7x=0$$

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.

$$-x^{2}-7x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -7 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-1\right)\times 8}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-\left(-7\right)±\sqrt{49+4\times 8}}{2\left(-1\right)}$$

Multiply 4 times 8.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2\left(-1\right)}$$

Add 49 to 32.

$$x=\frac{-\left(-7\right)±\sqrt{81}}{2\left(-1\right)}$$

Take the square root of 81.

$$x=\frac{-\left(-7\right)±9}{2\left(-1\right)}$$

The opposite of -7 is 7.

$$x=\frac{7±9}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{7±9}{-2}$$

Now solve the equation x=\frac{7±9}{-2} when ± is plus. Add 7 to 9.

$$x=\frac{16}{-2}$$

Divide 16 by -2.

$$x=-8$$

Now solve the equation x=\frac{7±9}{-2} when ± is minus. Subtract 9 from 7.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-8 x=1$$

Steps for Completing the Square

Consider \left(3x-1\right)\left(3x+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 1.

$$\left(3x\right)^{2}-1-\left(5x-4\right)\left(2x+3\right)=3$$

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-1-\left(5x-4\right)\left(2x+3\right)=3$$

Calculate 3 to the power of 2 and get 9.

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

Use the distributive property to multiply 5x-4 by 2x+3 and combine like terms.

$$9x^{2}-1-\left(10x^{2}+7x-12\right)=3$$

To find the opposite of 10x^{2}+7x-12, find the opposite of each term.

$$9x^{2}-1-10x^{2}-7x+12=3$$

Combine 9x^{2} and -10x^{2} to get -x^{2}.

$$-x^{2}-1-7x+12=3$$

Add -1 and 12 to get 11.

$$-x^{2}+11-7x=3$$

Subtract 11 from both sides.

$$-x^{2}-7x=3-11$$

Subtract 11 from 3 to get -8.

$$-x^{2}-7x=-8$$

Divide both sides by -1.

$$\frac{-x^{2}-7x}{-1}=-\frac{8}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\left(-\frac{7}{-1}\right)x=-\frac{8}{-1}$$

Divide -7 by -1.

$$x^{2}+7x=-\frac{8}{-1}$$

Divide -8 by -1.

$$x^{2}+7x=8$$

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

$$x^{2}+7x+\left(\frac{7}{2}\right)^{2}=8+\left(\frac{7}{2}\right)^{2}$$

Square \frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+7x+\frac{49}{4}=8+\frac{49}{4}$$

Add 8 to \frac{49}{4}.

$$x^{2}+7x+\frac{49}{4}=\frac{81}{4}$$

Factor x^{2}+7x+\frac{49}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x+\frac{7}{2}=\frac{9}{2} x+\frac{7}{2}=-\frac{9}{2}$$

Subtract \frac{7}{2} from both sides of the equation.

$$x=1 x=-8$$