Solve (x-2)^2=3x+12

$$( x - 2 ) ^ { 2 } = 3 x + 12$$

Solve for x

$$x=-1$$

$$x=8$$

Steps Using Factoring

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

$$x^{2}-4x+4=3x+12$$

Subtract 3x from both sides.

$$x^{2}-4x+4-3x=12$$

Combine -4x and -3x to get -7x.

$$x^{2}-7x+4=12$$

Subtract 12 from both sides.

$$x^{2}-7x+4-12=0$$

Subtract 12 from 4 to get -8.

$$x^{2}-7x-8=0$$

To solve the equation, factor x^{2}-7x-8 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-7 ab=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=-8 b=1$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-8\right)\left(x+1\right)$$

To find equation solutions, solve x-8=0 and x+1=0.

$$x=8 x=-1$$

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}.

$$x^{2}-4x+4=3x+12$$

Subtract 3x from both sides.

$$x^{2}-4x+4-3x=12$$

Combine -4x and -3x to get -7x.

$$x^{2}-7x+4=12$$

Subtract 12 from both sides.

$$x^{2}-7x+4-12=0$$

Subtract 12 from 4 to get -8.

$$x^{2}-7x-8=0$$

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

$$a+b=-7 ab=1\left(-8\right)=-8$$

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 -8.

$$1,-8 2,-4$$

Calculate the sum for each pair.

$$1-8=-7 2-4=-2$$

The solution is the pair that gives sum -7.

$$a=-8 b=1$$

Rewrite x^{2}-7x-8 as \left(x^{2}-8x\right)+\left(x-8\right).

$$\left(x^{2}-8x\right)+\left(x-8\right)$$

Factor out x in x^{2}-8x.

$$x\left(x-8\right)+x-8$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x+1\right)$$

To find equation solutions, solve x-8=0 and x+1=0.

$$x=8 x=-1$$

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}.

$$x^{2}-4x+4=3x+12$$

Subtract 3x from both sides.

$$x^{2}-4x+4-3x=12$$

Combine -4x and -3x to get -7x.

$$x^{2}-7x+4=12$$

Subtract 12 from both sides.

$$x^{2}-7x+4-12=0$$

Subtract 12 from 4 to get -8.

$$x^{2}-7x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\left(-8\right)}}{2}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\left(-8\right)}}{2}$$

Multiply -4 times -8.

$$x=\frac{-\left(-7\right)±\sqrt{49+32}}{2}$$

Add 49 to 32.

$$x=\frac{-\left(-7\right)±\sqrt{81}}{2}$$

Take the square root of 81.

$$x=\frac{-\left(-7\right)±9}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±9}{2}$$

Now solve the equation x=\frac{7±9}{2} when ± is plus. Add 7 to 9.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{7±9}{2} when ± is minus. Subtract 9 from 7.

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

Divide -2 by 2.

$$x=-1$$

The equation is now solved.

$$x=8 x=-1$$

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}.

$$x^{2}-4x+4=3x+12$$

Subtract 3x from both sides.

$$x^{2}-4x+4-3x=12$$

Combine -4x and -3x to get -7x.

$$x^{2}-7x+4=12$$

Subtract 4 from both sides.

$$x^{2}-7x=12-4$$

Subtract 4 from 12 to get 8.

$$x^{2}-7x=8$$

Divide -7, the coefficient of the x term, by 2 to get -\frac{7}{2}. Then add the square of -\frac{7}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-7x+\left(-\frac{7}{2}\right)^{2}=8+\left(-\frac{7}{2}\right)^{2}$$

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=8+\frac{49}{4}$$

Add 8 to \frac{49}{4}.

$$x^{2}-7x+\frac{49}{4}=\frac{81}{4}$$

Factor x^{2}-7x+\frac{49}{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}.

$$\left(x-\frac{7}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

$$x-\frac{7}{2}=\frac{9}{2} x-\frac{7}{2}=-\frac{9}{2}$$

Add \frac{7}{2} to both sides of the equation.

$$x=8 x=-1$$