Solve (-4x-2)^2-9=0

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

Solve for x

$$x = -\frac{5}{4} = -1\frac{1}{4} = -1.25$$

$$x=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4x-2\right)^{2}.

$$16x^{2}+16x+4-9=0$$

Subtract 9 from 4 to get -5.

$$16x^{2}+16x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=16 ab=16\left(-5\right)=-80$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -80.

$$-1,80 -2,40 -4,20 -5,16 -8,10$$

Calculate the sum for each pair.

$$-1+80=79 -2+40=38 -4+20=16 -5+16=11 -8+10=2$$

The solution is the pair that gives sum 16.

$$a=-4 b=20$$

Rewrite 16x^{2}+16x-5 as \left(16x^{2}-4x\right)+\left(20x-5\right).

$$\left(16x^{2}-4x\right)+\left(20x-5\right)$$

Factor out 4x in the first and 5 in the second group.

$$4x\left(4x-1\right)+5\left(4x-1\right)$$

Factor out common term 4x-1 by using distributive property.

$$\left(4x-1\right)\left(4x+5\right)$$

To find equation solutions, solve 4x-1=0 and 4x+5=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4x-2\right)^{2}.

$$16x^{2}+16x+4-9=0$$

Subtract 9 from 4 to get -5.

$$16x^{2}+16x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 16 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-16±\sqrt{16^{2}-4\times 16\left(-5\right)}}{2\times 16}$$

Square 16.

$$x=\frac{-16±\sqrt{256-4\times 16\left(-5\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-16±\sqrt{256-64\left(-5\right)}}{2\times 16}$$

Multiply -64 times -5.

$$x=\frac{-16±\sqrt{256+320}}{2\times 16}$$

Add 256 to 320.

$$x=\frac{-16±\sqrt{576}}{2\times 16}$$

Take the square root of 576.

$$x=\frac{-16±24}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-16±24}{32}$$

Now solve the equation x=\frac{-16±24}{32} when ± is plus. Add -16 to 24.

$$x=\frac{8}{32}$$

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

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

Now solve the equation x=\frac{-16±24}{32} when ± is minus. Subtract 24 from -16.

$$x=-\frac{40}{32}$$

Reduce the fraction \frac{-40}{32} to lowest terms by extracting and canceling out 8.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-4x-2\right)^{2}.

$$16x^{2}+16x+4-9=0$$

Subtract 9 from 4 to get -5.

$$16x^{2}+16x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$16x^{2}+16x=5$$

Divide both sides by 16.

$$\frac{16x^{2}+16x}{16}=\frac{5}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{16}{16}x=\frac{5}{16}$$

Divide 16 by 16.

$$x^{2}+x=\frac{5}{16}$$

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

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=\frac{5}{16}+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=\frac{5}{16}+\frac{1}{4}$$

Add \frac{5}{16} to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+x+\frac{1}{4}=\frac{9}{16}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{9}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{9}{16}}$$

Simplify.

$$x+\frac{1}{2}=\frac{3}{4} x+\frac{1}{2}=-\frac{3}{4}$$

Subtract \frac{1}{2} from both sides of the equation.

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