Solve (2x-3)(2x+3)=27

$$(2x-3)(2x+3)=27$$

Solve for x

$$x=-3$$

$$x=3$$

Steps by Finding Square Root

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.

$$\left(2x\right)^{2}-9=27$$

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

$$2^{2}x^{2}-9=27$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-9=27$$

Add 9 to both sides.

$$4x^{2}=27+9$$

Add 27 and 9 to get 36.

$$4x^{2}=36$$

Divide both sides by 4.

$$x^{2}=\frac{36}{4}$$

Divide 36 by 4 to get 9.

$$x^{2}=9$$

Take the square root of both sides of the equation.

$$x=3 x=-3$$

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.

$$\left(2x\right)^{2}-9=27$$

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

$$2^{2}x^{2}-9=27$$

Calculate 2 to the power of 2 and get 4.

$$4x^{2}-9=27$$

Subtract 27 from both sides.

$$4x^{2}-9-27=0$$

Subtract 27 from -9 to get -36.

$$4x^{2}-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, 0 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 4\left(-36\right)}}{2\times 4}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 4\left(-36\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{0±\sqrt{-16\left(-36\right)}}{2\times 4}$$

Multiply -16 times -36.

$$x=\frac{0±\sqrt{576}}{2\times 4}$$

Take the square root of 576.

$$x=\frac{0±24}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{0±24}{8}$$

Now solve the equation x=\frac{0±24}{8} when ± is plus. Divide 24 by 8.

$$x=3$$

Now solve the equation x=\frac{0±24}{8} when ± is minus. Divide -24 by 8.

$$x=-3$$

The equation is now solved.

$$x=3 x=-3$$