Solve 3(x-2)^2-7=20

$$3 ( x - 2 ) ^ { 2 } - 7 = 20$$

Solve for x

$$x=5$$

$$x=-1$$

Steps Using Factoring By Grouping

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

$$3\left(x^{2}-4x+4\right)-7=20$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}-12x+12-7=20$$

Subtract 7 from 12 to get 5.

$$3x^{2}-12x+5=20$$

Subtract 20 from both sides.

$$3x^{2}-12x+5-20=0$$

Subtract 20 from 5 to get -15.

$$3x^{2}-12x-15=0$$

Divide both sides by 3.

$$x^{2}-4x-5=0$$

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

$$a+b=-4 ab=1\left(-5\right)=-5$$

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.

$$a=-5 b=1$$

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

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

Factor out x in x^{2}-5x.

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

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

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

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

$$x=5 x=-1$$

Steps Using the Quadratic Formula

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

$$3\left(x^{2}-4x+4\right)-7=20$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}-12x+12-7=20$$

Subtract 7 from 12 to get 5.

$$3x^{2}-12x+5=20$$

Subtract 20 from both sides.

$$3x^{2}-12x+5-20=0$$

Subtract 20 from 5 to get -15.

$$3x^{2}-12x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -12 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 3\left(-15\right)}}{2\times 3}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 3\left(-15\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-12\right)±\sqrt{144-12\left(-15\right)}}{2\times 3}$$

Multiply -12 times -15.

$$x=\frac{-\left(-12\right)±\sqrt{144+180}}{2\times 3}$$

Add 144 to 180.

$$x=\frac{-\left(-12\right)±\sqrt{324}}{2\times 3}$$

Take the square root of 324.

$$x=\frac{-\left(-12\right)±18}{2\times 3}$$

The opposite of -12 is 12.

$$x=\frac{12±18}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{12±18}{6}$$

Now solve the equation x=\frac{12±18}{6} when ± is plus. Add 12 to 18.

$$x=\frac{30}{6}$$

Divide 30 by 6.

$$x=5$$

Now solve the equation x=\frac{12±18}{6} when ± is minus. Subtract 18 from 12.

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

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=5 x=-1$$

Steps for Completing the Square

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

$$3\left(x^{2}-4x+4\right)-7=20$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}-12x+12-7=20$$

Subtract 7 from 12 to get 5.

$$3x^{2}-12x+5=20$$

Subtract 5 from both sides.

$$3x^{2}-12x=20-5$$

Subtract 5 from 20 to get 15.

$$3x^{2}-12x=15$$

Divide both sides by 3.

$$\frac{3x^{2}-12x}{3}=\frac{15}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\left(-\frac{12}{3}\right)x=\frac{15}{3}$$

Divide -12 by 3.

$$x^{2}-4x=\frac{15}{3}$$

Divide 15 by 3.

$$x^{2}-4x=5$$

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

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

Square -2.

$$x^{2}-4x+4=5+4$$

Add 5 to 4.

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

Factor x^{2}-4x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-2=3 x-2=-3$$

Add 2 to both sides of the equation.

$$x=5 x=-1$$