Solve 12(x-1)^2+27=75

$$12 ( x - 1 ) ^ { 2 } + 27 = 75$$

Solve for x

$$x=3$$

$$x=-1$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$12\left(x^{2}-2x+1\right)+27=75$$

Use the distributive property to multiply 12 by x^{2}-2x+1.

$$12x^{2}-24x+12+27=75$$

Add 12 and 27 to get 39.

$$12x^{2}-24x+39=75$$

Subtract 75 from both sides.

$$12x^{2}-24x+39-75=0$$

Subtract 75 from 39 to get -36.

$$12x^{2}-24x-36=0$$

Divide both sides by 12.

$$x^{2}-2x-3=0$$

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

$$a+b=-2 ab=1\left(-3\right)=-3$$

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. The only such pair is the system solution.

$$a=-3 b=1$$

Rewrite x^{2}-2x-3 as \left(x^{2}-3x\right)+\left(x-3\right).

$$\left(x^{2}-3x\right)+\left(x-3\right)$$

Factor out x in x^{2}-3x.

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

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

$$\left(x-3\right)\left(x+1\right)$$

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

$$x=3 x=-1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$12\left(x^{2}-2x+1\right)+27=75$$

Use the distributive property to multiply 12 by x^{2}-2x+1.

$$12x^{2}-24x+12+27=75$$

Add 12 and 27 to get 39.

$$12x^{2}-24x+39=75$$

Subtract 75 from both sides.

$$12x^{2}-24x+39-75=0$$

Subtract 75 from 39 to get -36.

$$12x^{2}-24x-36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 12 for a, -24 for b, and -36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 12\left(-36\right)}}{2\times 12}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 12\left(-36\right)}}{2\times 12}$$

Multiply -4 times 12.

$$x=\frac{-\left(-24\right)±\sqrt{576-48\left(-36\right)}}{2\times 12}$$

Multiply -48 times -36.

$$x=\frac{-\left(-24\right)±\sqrt{576+1728}}{2\times 12}$$

Add 576 to 1728.

$$x=\frac{-\left(-24\right)±\sqrt{2304}}{2\times 12}$$

Take the square root of 2304.

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

The opposite of -24 is 24.

$$x=\frac{24±48}{2\times 12}$$

Multiply 2 times 12.

$$x=\frac{24±48}{24}$$

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

$$x=\frac{72}{24}$$

Divide 72 by 24.

$$x=3$$

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

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

Divide -24 by 24.

$$x=-1$$

The equation is now solved.

$$x=3 x=-1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-1\right)^{2}.

$$12\left(x^{2}-2x+1\right)+27=75$$

Use the distributive property to multiply 12 by x^{2}-2x+1.

$$12x^{2}-24x+12+27=75$$

Add 12 and 27 to get 39.

$$12x^{2}-24x+39=75$$

Subtract 39 from both sides.

$$12x^{2}-24x=75-39$$

Subtract 39 from 75 to get 36.

$$12x^{2}-24x=36$$

Divide both sides by 12.

$$\frac{12x^{2}-24x}{12}=\frac{36}{12}$$

Dividing by 12 undoes the multiplication by 12.

$$x^{2}+\left(-\frac{24}{12}\right)x=\frac{36}{12}$$

Divide -24 by 12.

$$x^{2}-2x=\frac{36}{12}$$

Divide 36 by 12.

$$x^{2}-2x=3$$

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

$$x^{2}-2x+1=3+1$$

Add 3 to 1.

$$x^{2}-2x+1=4$$

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

$$\left(x-1\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-1=2 x-1=-2$$

Add 1 to both sides of the equation.

$$x=3 x=-1$$