Solve (2x+4)*(2x-4)=4x-(3*2x)

$$( 2 x + 4 ) \cdot ( 2 x - 4 ) = 4 x - ( 3 \cdot 2 x )$$

Solve for x

$$x = \frac{\sqrt{65} - 1}{4} \approx 1.765564437$$

$$x=\frac{-\sqrt{65}-1}{4}\approx -2.265564437$$

Steps Using the Quadratic Formula

Consider \left(2x+4\right)\left(2x-4\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 4.

$$\left(2x\right)^{2}-16=4x-3\times 2x$$

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

$$2^{2}x^{2}-16=4x-3\times 2x$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-16=4x-3\times 2x$$

Multiply 3 and 2 to get 6.

$$4x^{2}-16=4x-6x$$

Combine 4x and -6x to get -2x.

$$4x^{2}-16=-2x$$

Add 2x to both sides.

$$4x^{2}-16+2x=0$$

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.

$$4x^{2}+2x-16=0$$

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

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

Square 2.

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

Multiply -4 times 4.

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

Multiply -16 times -16.

$$x=\frac{-2±\sqrt{4+256}}{2\times 4}$$

Add 4 to 256.

$$x=\frac{-2±\sqrt{260}}{2\times 4}$$

Take the square root of 260.

$$x=\frac{-2±2\sqrt{65}}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{-2±2\sqrt{65}}{8}$$

Now solve the equation x=\frac{-2±2\sqrt{65}}{8} when ± is plus. Add -2 to 2\sqrt{65}.

$$x=\frac{2\sqrt{65}-2}{8}$$

Divide -2+2\sqrt{65} by 8.

$$x=\frac{\sqrt{65}-1}{4}$$

Now solve the equation x=\frac{-2±2\sqrt{65}}{8} when ± is minus. Subtract 2\sqrt{65} from -2.

$$x=\frac{-2\sqrt{65}-2}{8}$$

Divide -2-2\sqrt{65} by 8.

$$x=\frac{-\sqrt{65}-1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{65}-1}{4} x=\frac{-\sqrt{65}-1}{4}$$

Steps for Completing the Square

Consider \left(2x+4\right)\left(2x-4\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 4.

$$\left(2x\right)^{2}-16=4x-3\times 2x$$

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

$$2^{2}x^{2}-16=4x-3\times 2x$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-16=4x-3\times 2x$$

Multiply 3 and 2 to get 6.

$$4x^{2}-16=4x-6x$$

Combine 4x and -6x to get -2x.

$$4x^{2}-16=-2x$$

Add 2x to both sides.

$$4x^{2}-16+2x=0$$

Add 16 to both sides. Anything plus zero gives itself.

$$4x^{2}+2x=16$$

Divide both sides by 4.

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

Dividing by 4 undoes the multiplication by 4.

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

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

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

Divide 16 by 4.

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

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.

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

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

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

Add 4 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{65}{16}$$

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}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{65}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{65}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{65}}{4} x+\frac{1}{4}=-\frac{\sqrt{65}}{4}$$

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

$$x=\frac{\sqrt{65}-1}{4} x=\frac{-\sqrt{65}-1}{4}$$