Solve (4x-3)^2=36

$$( 4 x - 3 ) ^ { 2 } = 36$$

Solve for x

$$x = \frac{9}{4} = 2\frac{1}{4} = 2.25$$

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

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}.

$$16x^{2}-24x+9=36$$

Subtract 36 from both sides.

$$16x^{2}-24x+9-36=0$$

Subtract 36 from 9 to get -27.

$$16x^{2}-24x-27=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-27. To find a and b, set up a system to be solved.

$$a+b=-24 ab=16\left(-27\right)=-432$$

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 -432.

$$1,-432 2,-216 3,-144 4,-108 6,-72 8,-54 9,-48 12,-36 16,-27 18,-24$$

Calculate the sum for each pair.

$$1-432=-431 2-216=-214 3-144=-141 4-108=-104 6-72=-66 8-54=-46 9-48=-39 12-36=-24 16-27=-11 18-24=-6$$

The solution is the pair that gives sum -24.

$$a=-36 b=12$$

Rewrite 16x^{2}-24x-27 as \left(16x^{2}-36x\right)+\left(12x-27\right).

$$\left(16x^{2}-36x\right)+\left(12x-27\right)$$

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

$$4x\left(4x-9\right)+3\left(4x-9\right)$$

Factor out common term 4x-9 by using distributive property.

$$\left(4x-9\right)\left(4x+3\right)$$

To find equation solutions, solve 4x-9=0 and 4x+3=0.

$$x=\frac{9}{4} x=-\frac{3}{4}$$

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}.

$$16x^{2}-24x+9=36$$

Subtract 36 from both sides.

$$16x^{2}-24x+9-36=0$$

Subtract 36 from 9 to get -27.

$$16x^{2}-24x-27=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -24 for b, and -27 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 16\left(-27\right)}}{2\times 16}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 16\left(-27\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-24\right)±\sqrt{576-64\left(-27\right)}}{2\times 16}$$

Multiply -64 times -27.

$$x=\frac{-\left(-24\right)±\sqrt{576+1728}}{2\times 16}$$

Add 576 to 1728.

$$x=\frac{-\left(-24\right)±\sqrt{2304}}{2\times 16}$$

Take the square root of 2304.

$$x=\frac{-\left(-24\right)±48}{2\times 16}$$

The opposite of -24 is 24.

$$x=\frac{24±48}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{24±48}{32}$$

Now solve the equation x=\frac{24±48}{32} when ± is plus. Add 24 to 48.

$$x=\frac{72}{32}$$

Reduce the fraction \frac{72}{32} to lowest terms by extracting and canceling out 8.

$$x=\frac{9}{4}$$

Now solve the equation x=\frac{24±48}{32} when ± is minus. Subtract 48 from 24.

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

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

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

The equation is now solved.

$$x=\frac{9}{4} x=-\frac{3}{4}$$

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}.

$$16x^{2}-24x+9=36$$

Subtract 9 from both sides.

$$16x^{2}-24x=36-9$$

Subtract 9 from 36 to get 27.

$$16x^{2}-24x=27$$

Divide both sides by 16.

$$\frac{16x^{2}-24x}{16}=\frac{27}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{24}{16}\right)x=\frac{27}{16}$$

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

$$x^{2}-\frac{3}{2}x=\frac{27}{16}$$

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

$$x^{2}-\frac{3}{2}x+\left(-\frac{3}{4}\right)^{2}=\frac{27}{16}+\left(-\frac{3}{4}\right)^{2}$$

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

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{27+9}{16}$$

Add \frac{27}{16} to \frac{9}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{2}x+\frac{9}{16}=\frac{9}{4}$$

Factor x^{2}-\frac{3}{2}x+\frac{9}{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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{4}\right)^{2}}=\sqrt{\frac{9}{4}}$$

Simplify.

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

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

$$x=\frac{9}{4} x=-\frac{3}{4}$$