Solve (4x^2-9)+(2x-3)(4x+5)=0

$$( 4 x ^ { 2 } - 9 ) + ( 2 x - 3 ) ( 4 x + 5 ) = 0$$

Solve for x

$$x = -\frac{4}{3} = -1\frac{1}{3} \approx -1.333333333$$

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

Steps Using Factoring By Grouping

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

$$4x^{2}-9+8x^{2}-2x-15=0$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$12x^{2}-9-2x-15=0$$

Subtract 15 from -9 to get -24.

$$12x^{2}-24-2x=0$$

Divide both sides by 2.

$$6x^{2}-12-x=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$6x^{2}-x-12=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 6x^{2}+ax+bx-12. To find a and b, set up a system to be solved.

$$a+b=-1 ab=6\left(-12\right)=-72$$

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 -72.

$$1,-72 2,-36 3,-24 4,-18 6,-12 8,-9$$

Calculate the sum for each pair.

$$1-72=-71 2-36=-34 3-24=-21 4-18=-14 6-12=-6 8-9=-1$$

The solution is the pair that gives sum -1.

$$a=-9 b=8$$

Rewrite 6x^{2}-x-12 as \left(6x^{2}-9x\right)+\left(8x-12\right).

$$\left(6x^{2}-9x\right)+\left(8x-12\right)$$

Factor out 3x in the first and 4 in the second group.

$$3x\left(2x-3\right)+4\left(2x-3\right)$$

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

$$\left(2x-3\right)\left(3x+4\right)$$

To find equation solutions, solve 2x-3=0 and 3x+4=0.

$$x=\frac{3}{2} x=-\frac{4}{3}$$

Steps Using the Quadratic Formula

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

$$4x^{2}-9+8x^{2}-2x-15=0$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$12x^{2}-9-2x-15=0$$

Subtract 15 from -9 to get -24.

$$12x^{2}-24-2x=0$$

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.

$$12x^{2}-2x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -2 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 12\left(-24\right)}}{2\times 12}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 12\left(-24\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-2\right)±\sqrt{4-48\left(-24\right)}}{2\times 12}$$

Multiply -48 times -24.

$$x=\frac{-\left(-2\right)±\sqrt{4+1152}}{2\times 12}$$

Add 4 to 1152.

$$x=\frac{-\left(-2\right)±\sqrt{1156}}{2\times 12}$$

Take the square root of 1156.

$$x=\frac{-\left(-2\right)±34}{2\times 12}$$

The opposite of -2 is 2.

$$x=\frac{2±34}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{2±34}{24}$$

Now solve the equation x=\frac{2±34}{24} when ± is plus. Add 2 to 34.

$$x=\frac{36}{24}$$

Reduce the fraction \frac{36}{24} to lowest terms by extracting and canceling out 12.

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

Now solve the equation x=\frac{2±34}{24} when ± is minus. Subtract 34 from 2.

$$x=-\frac{32}{24}$$

Reduce the fraction \frac{-32}{24} to lowest terms by extracting and canceling out 8.

$$x=-\frac{4}{3}$$

The equation is now solved.

$$x=\frac{3}{2} x=-\frac{4}{3}$$

Steps for Completing the Square

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

$$4x^{2}-9+8x^{2}-2x-15=0$$

Combine 4x^{2} and 8x^{2} to get 12x^{2}.

$$12x^{2}-9-2x-15=0$$

Subtract 15 from -9 to get -24.

$$12x^{2}-24-2x=0$$

Add 24 to both sides. Anything plus zero gives itself.

$$12x^{2}-2x=24$$

Divide both sides by 12.

$$\frac{12x^{2}-2x}{12}=\frac{24}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{2}{12}\right)x=\frac{24}{12}$$

Reduce the fraction \frac{-2}{12} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{6}x=\frac{24}{12}$$

Divide 24 by 12.

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

Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=2+\left(-\frac{1}{12}\right)^{2}$$

Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=2+\frac{1}{144}$$

Add 2 to \frac{1}{144}.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{289}{144}$$

Factor x^{2}-\frac{1}{6}x+\frac{1}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{12}\right)^{2}=\frac{289}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{12}\right)^{2}}=\sqrt{\frac{289}{144}}$$

Simplify.

$$x-\frac{1}{12}=\frac{17}{12} x-\frac{1}{12}=-\frac{17}{12}$$

Add \frac{1}{12} to both sides of the equation.

$$x=\frac{3}{2} x=-\frac{4}{3}$$