Solve 2(x-5)^2=4(x+7)

$$2 ( x - 5 ) ^ { 2 } = 4 ( x + 7 )$$

Solve for x

$$x=1$$

$$x=11$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$2\left(x^{2}-10x+25\right)=4\left(x+7\right)$$

Use the distributive property to multiply 2 by x^{2}-10x+25.

$$2x^{2}-20x+50=4\left(x+7\right)$$

Use the distributive property to multiply 4 by x+7.

$$2x^{2}-20x+50=4x+28$$

Subtract 4x from both sides.

$$2x^{2}-20x+50-4x=28$$

Combine -20x and -4x to get -24x.

$$2x^{2}-24x+50=28$$

Subtract 28 from both sides.

$$2x^{2}-24x+50-28=0$$

Subtract 28 from 50 to get 22.

$$2x^{2}-24x+22=0$$

Divide both sides by 2.

$$x^{2}-12x+11=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+11. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 11=11$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. The only such pair is the system solution.

$$a=-11 b=-1$$

Rewrite x^{2}-12x+11 as \left(x^{2}-11x\right)+\left(-x+11\right).

$$\left(x^{2}-11x\right)+\left(-x+11\right)$$

Factor out x in the first and -1 in the second group.

$$x\left(x-11\right)-\left(x-11\right)$$

Factor out common term x-11 by using distributive property.

$$\left(x-11\right)\left(x-1\right)$$

To find equation solutions, solve x-11=0 and x-1=0.

$$x=11 x=1$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$2\left(x^{2}-10x+25\right)=4\left(x+7\right)$$

Use the distributive property to multiply 2 by x^{2}-10x+25.

$$2x^{2}-20x+50=4\left(x+7\right)$$

Use the distributive property to multiply 4 by x+7.

$$2x^{2}-20x+50=4x+28$$

Subtract 4x from both sides.

$$2x^{2}-20x+50-4x=28$$

Combine -20x and -4x to get -24x.

$$2x^{2}-24x+50=28$$

Subtract 28 from both sides.

$$2x^{2}-24x+50-28=0$$

Subtract 28 from 50 to get 22.

$$2x^{2}-24x+22=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -24 for b, and 22 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 2\times 22}}{2\times 2}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 2\times 22}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-24\right)±\sqrt{576-8\times 22}}{2\times 2}$$

Multiply -8 times 22.

$$x=\frac{-\left(-24\right)±\sqrt{576-176}}{2\times 2}$$

Add 576 to -176.

$$x=\frac{-\left(-24\right)±\sqrt{400}}{2\times 2}$$

Take the square root of 400.

$$x=\frac{-\left(-24\right)±20}{2\times 2}$$

The opposite of -24 is 24.

$$x=\frac{24±20}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{24±20}{4}$$

Now solve the equation x=\frac{24±20}{4} when ± is plus. Add 24 to 20.

$$x=\frac{44}{4}$$

Divide 44 by 4.

$$x=11$$

Now solve the equation x=\frac{24±20}{4} when ± is minus. Subtract 20 from 24.

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

Divide 4 by 4.

$$x=1$$

The equation is now solved.

$$x=11 x=1$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$2\left(x^{2}-10x+25\right)=4\left(x+7\right)$$

Use the distributive property to multiply 2 by x^{2}-10x+25.

$$2x^{2}-20x+50=4\left(x+7\right)$$

Use the distributive property to multiply 4 by x+7.

$$2x^{2}-20x+50=4x+28$$

Subtract 4x from both sides.

$$2x^{2}-20x+50-4x=28$$

Combine -20x and -4x to get -24x.

$$2x^{2}-24x+50=28$$

Subtract 50 from both sides.

$$2x^{2}-24x=28-50$$

Subtract 50 from 28 to get -22.

$$2x^{2}-24x=-22$$

Divide both sides by 2.

$$\frac{2x^{2}-24x}{2}=-\frac{22}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{24}{2}\right)x=-\frac{22}{2}$$

Divide -24 by 2.

$$x^{2}-12x=-\frac{22}{2}$$

Divide -22 by 2.

$$x^{2}-12x=-11$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-11+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-11+36$$

Add -11 to 36.

$$x^{2}-12x+36=25$$

Factor x^{2}-12x+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-6\right)^{2}=25$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{25}$$

Simplify.

$$x-6=5 x-6=-5$$

Add 6 to both sides of the equation.

$$x=11 x=1$$