Solve 42=4left(x-2right)^2+6

$$42 = 4 { \left( x-2 \right) }^{ 2 } +6$$

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}.

$$42=4\left(x^{2}-4x+4\right)+6$$

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

$$42=4x^{2}-16x+16+6$$

Add 16 and 6 to get 22.

$$42=4x^{2}-16x+22$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-16x+22=42$$

Subtract 42 from both sides.

$$4x^{2}-16x+22-42=0$$

Subtract 42 from 22 to get -20.

$$4x^{2}-16x-20=0$$

Divide both sides by 4.

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

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}.

$$42=4\left(x^{2}-4x+4\right)+6$$

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

$$42=4x^{2}-16x+16+6$$

Add 16 and 6 to get 22.

$$42=4x^{2}-16x+22$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-16x+22=42$$

Subtract 42 from both sides.

$$4x^{2}-16x+22-42=0$$

Subtract 42 from 22 to get -20.

$$4x^{2}-16x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -16 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\times 4\left(-20\right)}}{2\times 4}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\times 4\left(-20\right)}}{2\times 4}$$

Multiply -4 times 4.

$$x=\frac{-\left(-16\right)±\sqrt{256-16\left(-20\right)}}{2\times 4}$$

Multiply -16 times -20.

$$x=\frac{-\left(-16\right)±\sqrt{256+320}}{2\times 4}$$

Add 256 to 320.

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

Take the square root of 576.

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

The opposite of -16 is 16.

$$x=\frac{16±24}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{16±24}{8}$$

Now solve the equation x=\frac{16±24}{8} when ± is plus. Add 16 to 24.

$$x=\frac{40}{8}$$

Divide 40 by 8.

$$x=5$$

Now solve the equation x=\frac{16±24}{8} when ± is minus. Subtract 24 from 16.

$$x=-\frac{8}{8}$$

Divide -8 by 8.

$$x=-1$$

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}.

$$42=4\left(x^{2}-4x+4\right)+6$$

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

$$42=4x^{2}-16x+16+6$$

Add 16 and 6 to get 22.

$$42=4x^{2}-16x+22$$

Swap sides so that all variable terms are on the left hand side.

$$4x^{2}-16x+22=42$$

Subtract 22 from both sides.

$$4x^{2}-16x=42-22$$

Subtract 22 from 42 to get 20.

$$4x^{2}-16x=20$$

Divide both sides by 4.

$$\frac{4x^{2}-16x}{4}=\frac{20}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}+\left(-\frac{16}{4}\right)x=\frac{20}{4}$$

Divide -16 by 4.

$$x^{2}-4x=\frac{20}{4}$$

Divide 20 by 4.

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

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.

$$x=5 x=-1$$