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

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

Solve for x

$$x=\frac{1}{6}\approx 0.166666667$$

$$x = \frac{5}{2} = 2\frac{1}{2} = 2.5$$

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}.

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

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

$$16x^{2}-24x+9=4x^{2}+8x+4$$

Subtract 4x^{2} from both sides.

$$16x^{2}-24x+9-4x^{2}=8x+4$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}-24x+9=8x+4$$

Subtract 8x from both sides.

$$12x^{2}-24x+9-8x=4$$

Combine -24x and -8x to get -32x.

$$12x^{2}-32x+9=4$$

Subtract 4 from both sides.

$$12x^{2}-32x+9-4=0$$

Subtract 4 from 9 to get 5.

$$12x^{2}-32x+5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 12x^{2}+ax+bx+5. To find a and b, set up a system to be solved.

$$a+b=-32 ab=12\times 5=60$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 60.

$$-1,-60 -2,-30 -3,-20 -4,-15 -5,-12 -6,-10$$

Calculate the sum for each pair.

$$-1-60=-61 -2-30=-32 -3-20=-23 -4-15=-19 -5-12=-17 -6-10=-16$$

The solution is the pair that gives sum -32.

$$a=-30 b=-2$$

Rewrite 12x^{2}-32x+5 as \left(12x^{2}-30x\right)+\left(-2x+5\right).

$$\left(12x^{2}-30x\right)+\left(-2x+5\right)$$

Factor out 6x in the first and -1 in the second group.

$$6x\left(2x-5\right)-\left(2x-5\right)$$

Factor out common term 2x-5 by using distributive property.

$$\left(2x-5\right)\left(6x-1\right)$$

To find equation solutions, solve 2x-5=0 and 6x-1=0.

$$x=\frac{5}{2} x=\frac{1}{6}$$

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}.

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

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

$$16x^{2}-24x+9=4x^{2}+8x+4$$

Subtract 4x^{2} from both sides.

$$16x^{2}-24x+9-4x^{2}=8x+4$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}-24x+9=8x+4$$

Subtract 8x from both sides.

$$12x^{2}-24x+9-8x=4$$

Combine -24x and -8x to get -32x.

$$12x^{2}-32x+9=4$$

Subtract 4 from both sides.

$$12x^{2}-32x+9-4=0$$

Subtract 4 from 9 to get 5.

$$12x^{2}-32x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -32 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 12\times 5}}{2\times 12}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 12\times 5}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-32\right)±\sqrt{1024-48\times 5}}{2\times 12}$$

Multiply -48 times 5.

$$x=\frac{-\left(-32\right)±\sqrt{1024-240}}{2\times 12}$$

Add 1024 to -240.

$$x=\frac{-\left(-32\right)±\sqrt{784}}{2\times 12}$$

Take the square root of 784.

$$x=\frac{-\left(-32\right)±28}{2\times 12}$$

The opposite of -32 is 32.

$$x=\frac{32±28}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{32±28}{24}$$

Now solve the equation x=\frac{32±28}{24} when ± is plus. Add 32 to 28.

$$x=\frac{60}{24}$$

Reduce the fraction \frac{60}{24} to lowest terms by extracting and canceling out 12.

$$x=\frac{5}{2}$$

Now solve the equation x=\frac{32±28}{24} when ± is minus. Subtract 28 from 32.

$$x=\frac{4}{24}$$

Reduce the fraction \frac{4}{24} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{6}$$

The equation is now solved.

$$x=\frac{5}{2} x=\frac{1}{6}$$

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}.

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

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

$$16x^{2}-24x+9=4x^{2}+8x+4$$

Subtract 4x^{2} from both sides.

$$16x^{2}-24x+9-4x^{2}=8x+4$$

Combine 16x^{2} and -4x^{2} to get 12x^{2}.

$$12x^{2}-24x+9=8x+4$$

Subtract 8x from both sides.

$$12x^{2}-24x+9-8x=4$$

Combine -24x and -8x to get -32x.

$$12x^{2}-32x+9=4$$

Subtract 9 from both sides.

$$12x^{2}-32x=4-9$$

Subtract 9 from 4 to get -5.

$$12x^{2}-32x=-5$$

Divide both sides by 12.

$$\frac{12x^{2}-32x}{12}=-\frac{5}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{32}{12}\right)x=-\frac{5}{12}$$

Reduce the fraction \frac{-32}{12} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{8}{3}x=-\frac{5}{12}$$

Divide -\frac{8}{3}, the coefficient of the x term, by 2 to get -\frac{4}{3}. Then add the square of -\frac{4}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{8}{3}x+\left(-\frac{4}{3}\right)^{2}=-\frac{5}{12}+\left(-\frac{4}{3}\right)^{2}$$

Square -\frac{4}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=-\frac{5}{12}+\frac{16}{9}$$

Add -\frac{5}{12} to \frac{16}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{8}{3}x+\frac{16}{9}=\frac{49}{36}$$

Factor x^{2}-\frac{8}{3}x+\frac{16}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{4}{3}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{4}{3}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{4}{3}=\frac{7}{6} x-\frac{4}{3}=-\frac{7}{6}$$

Add \frac{4}{3} to both sides of the equation.

$$x=\frac{5}{2} x=\frac{1}{6}$$