Solve 2x-3=(2x-3)(2x+3)

$$2x-3=(2x-3)(2x+3)$$

Solve for x

$$x=-1$$

$$x = \frac{3}{2} = 1\frac{1}{2} = 1.5$$

Steps Using the Quadratic Formula

Consider \left(2x-3\right)\left(2x+3\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 3.

$$2x-3=\left(2x\right)^{2}-9$$

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

$$2x-3=2^{2}x^{2}-9$$

Calculate 2 to the power of 2 and get 4.

$$2x-3=4x^{2}-9$$

Subtract 4x^{2} from both sides.

$$2x-3-4x^{2}=-9$$

Add 9 to both sides.

$$2x-3-4x^{2}+9=0$$

Add -3 and 9 to get 6.

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

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.

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

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

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

Square 2.

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

Multiply -4 times -4.

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

Multiply 16 times 6.

$$x=\frac{-2±\sqrt{4+96}}{2\left(-4\right)}$$

Add 4 to 96.

$$x=\frac{-2±\sqrt{100}}{2\left(-4\right)}$$

Take the square root of 100.

$$x=\frac{-2±10}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{-2±10}{-8}$$

Now solve the equation x=\frac{-2±10}{-8} when ± is plus. Add -2 to 10.

$$x=\frac{8}{-8}$$

Divide 8 by -8.

$$x=-1$$

Now solve the equation x=\frac{-2±10}{-8} when ± is minus. Subtract 10 from -2.

$$x=-\frac{12}{-8}$$

Reduce the fraction \frac{-12}{-8} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Consider \left(2x-3\right)\left(2x+3\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 3.

$$2x-3=\left(2x\right)^{2}-9$$

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

$$2x-3=2^{2}x^{2}-9$$

Calculate 2 to the power of 2 and get 4.

$$2x-3=4x^{2}-9$$

Subtract 4x^{2} from both sides.

$$2x-3-4x^{2}=-9$$

Add 3 to both sides.

$$2x-4x^{2}=-9+3$$

Add -9 and 3 to get -6.

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

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

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

Divide both sides by -4.

$$\frac{-4x^{2}+2x}{-4}=-\frac{6}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\frac{2}{-4}x=-\frac{6}{-4}$$

Reduce the fraction \frac{2}{-4} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{2}x=-\frac{6}{-4}$$

Reduce the fraction \frac{-6}{-4} to lowest terms by extracting and canceling out 2.

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

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.

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

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

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

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{5}{4} x-\frac{1}{4}=-\frac{5}{4}$$

Add \frac{1}{4} to both sides of the equation.

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