Solve left(2x+1right)left(2x-3right)+4x^2-9=0

$$\left( 2x+1 \right) \left( 2x-3 \right) +4 { x }^{ 2 } -9 = 0$$

Solve for x

$$x=-1$$

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

Steps Using Factoring By Grouping

Use the distributive property to multiply 2x+1 by 2x-3 and combine like terms.

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}-4x-3-9=0$$

Subtract 9 from -3 to get -12.

$$8x^{2}-4x-12=0$$

Divide both sides by 4.

$$2x^{2}-x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-1 ab=2\left(-3\right)=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=-3 b=2$$

Rewrite 2x^{2}-x-3 as \left(2x^{2}-3x\right)+\left(2x-3\right).

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

Factor out x in 2x^{2}-3x.

$$x\left(2x-3\right)+2x-3$$

Factor out common term 2x-3 by using distributive property.

$$\left(2x-3\right)\left(x+1\right)$$

To find equation solutions, solve 2x-3=0 and x+1=0.

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

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x+1 by 2x-3 and combine like terms.

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}-4x-3-9=0$$

Subtract 9 from -3 to get -12.

$$8x^{2}-4x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -4 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-4\right)±\sqrt{\left(-4\right)^{2}-4\times 8\left(-12\right)}}{2\times 8}$$

Square -4.

$$x=\frac{-\left(-4\right)±\sqrt{16-4\times 8\left(-12\right)}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-4\right)±\sqrt{16-32\left(-12\right)}}{2\times 8}$$

Multiply -32 times -12.

$$x=\frac{-\left(-4\right)±\sqrt{16+384}}{2\times 8}$$

Add 16 to 384.

$$x=\frac{-\left(-4\right)±\sqrt{400}}{2\times 8}$$

Take the square root of 400.

$$x=\frac{-\left(-4\right)±20}{2\times 8}$$

The opposite of -4 is 4.

$$x=\frac{4±20}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{4±20}{16}$$

Now solve the equation x=\frac{4±20}{16} when ± is plus. Add 4 to 20.

$$x=\frac{24}{16}$$

Reduce the fraction \frac{24}{16} to lowest terms by extracting and canceling out 8.

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

Now solve the equation x=\frac{4±20}{16} when ± is minus. Subtract 20 from 4.

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

Divide -16 by 16.

$$x=-1$$

The equation is now solved.

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

Steps for Completing the Square

Use the distributive property to multiply 2x+1 by 2x-3 and combine like terms.

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

Combine 4x^{2} and 4x^{2} to get 8x^{2}.

$$8x^{2}-4x-3-9=0$$

Subtract 9 from -3 to get -12.

$$8x^{2}-4x-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$8x^{2}-4x=12$$

Divide both sides by 8.

$$\frac{8x^{2}-4x}{8}=\frac{12}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{4}{8}\right)x=\frac{12}{8}$$

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

$$x^{2}-\frac{1}{2}x=\frac{12}{8}$$

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

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

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.

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