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

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=-\frac{3}{4}=-0.75$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9-\left(4x+3\right)=0$$

To find the opposite of 4x+3, find the opposite of each term.

$$16x^{2}+24x+9-4x-3=0$$

Combine 24x and -4x to get 20x.

$$16x^{2}+20x+9-3=0$$

Subtract 3 from 9 to get 6.

$$16x^{2}+20x+6=0$$

Divide both sides by 2.

$$8x^{2}+10x+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+3. To find a and b, set up a system to be solved.

$$a+b=10 ab=8\times 3=24$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 24.

$$1,24 2,12 3,8 4,6$$

Calculate the sum for each pair.

$$1+24=25 2+12=14 3+8=11 4+6=10$$

The solution is the pair that gives sum 10.

$$a=4 b=6$$

Rewrite 8x^{2}+10x+3 as \left(8x^{2}+4x\right)+\left(6x+3\right).

$$\left(8x^{2}+4x\right)+\left(6x+3\right)$$

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

$$4x\left(2x+1\right)+3\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(4x+3\right)$$

To find equation solutions, solve 2x+1=0 and 4x+3=0.

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9-\left(4x+3\right)=0$$

To find the opposite of 4x+3, find the opposite of each term.

$$16x^{2}+24x+9-4x-3=0$$

Combine 24x and -4x to get 20x.

$$16x^{2}+20x+9-3=0$$

Subtract 3 from 9 to get 6.

$$16x^{2}+20x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, 20 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\times 16\times 6}}{2\times 16}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\times 16\times 6}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-20±\sqrt{400-64\times 6}}{2\times 16}$$

Multiply -64 times 6.

$$x=\frac{-20±\sqrt{400-384}}{2\times 16}$$

Add 400 to -384.

$$x=\frac{-20±\sqrt{16}}{2\times 16}$$

Take the square root of 16.

$$x=\frac{-20±4}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{-20±4}{32}$$

Now solve the equation x=\frac{-20±4}{32} when ± is plus. Add -20 to 4.

$$x=-\frac{16}{32}$$

Reduce the fraction \frac{-16}{32} to lowest terms by extracting and canceling out 16.

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

Now solve the equation x=\frac{-20±4}{32} when ± is minus. Subtract 4 from -20.

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

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

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(4x+3\right)^{2}.

$$16x^{2}+24x+9-\left(4x+3\right)=0$$

To find the opposite of 4x+3, find the opposite of each term.

$$16x^{2}+24x+9-4x-3=0$$

Combine 24x and -4x to get 20x.

$$16x^{2}+20x+9-3=0$$

Subtract 3 from 9 to get 6.

$$16x^{2}+20x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$16x^{2}+20x=-6$$

Divide both sides by 16.

$$\frac{16x^{2}+20x}{16}=-\frac{6}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\frac{20}{16}x=-\frac{6}{16}$$

Reduce the fraction \frac{20}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{5}{4}x=-\frac{6}{16}$$

Reduce the fraction \frac{-6}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{5}{4}x=-\frac{3}{8}$$

Divide \frac{5}{4}, the coefficient of the x term, by 2 to get \frac{5}{8}. Then add the square of \frac{5}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{4}x+\left(\frac{5}{8}\right)^{2}=-\frac{3}{8}+\left(\frac{5}{8}\right)^{2}$$

Square \frac{5}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=-\frac{3}{8}+\frac{25}{64}$$

Add -\frac{3}{8} to \frac{25}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{4}x+\frac{25}{64}=\frac{1}{64}$$

Factor x^{2}+\frac{5}{4}x+\frac{25}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{8}\right)^{2}=\frac{1}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{8}\right)^{2}}=\sqrt{\frac{1}{64}}$$

Simplify.

$$x+\frac{5}{8}=\frac{1}{8} x+\frac{5}{8}=-\frac{1}{8}$$

Subtract \frac{5}{8} from both sides of the equation.

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