Solve left(x-2right)^2+left(x-2right)^2=7^2

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

Solve for x

$$x=\frac{7\sqrt{2}}{2}+2\approx 6.949747468$$

$$x=-\frac{7\sqrt{2}}{2}+2\approx -2.949747468$$

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}.

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

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Combine -4x and -4x to get -8x.

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

Add 4 and 4 to get 8.

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

Calculate 7 to the power of 2 and get 49.

$$2x^{2}-8x+8=49$$

Subtract 49 from both sides.

$$2x^{2}-8x+8-49=0$$

Subtract 49 from 8 to get -41.

$$2x^{2}-8x-41=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -8 for b, and -41 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\times 2\left(-41\right)}}{2\times 2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\times 2\left(-41\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-8\right)±\sqrt{64-8\left(-41\right)}}{2\times 2}$$

Multiply -8 times -41.

$$x=\frac{-\left(-8\right)±\sqrt{64+328}}{2\times 2}$$

Add 64 to 328.

$$x=\frac{-\left(-8\right)±\sqrt{392}}{2\times 2}$$

Take the square root of 392.

$$x=\frac{-\left(-8\right)±14\sqrt{2}}{2\times 2}$$

The opposite of -8 is 8.

$$x=\frac{8±14\sqrt{2}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{8±14\sqrt{2}}{4}$$

Now solve the equation x=\frac{8±14\sqrt{2}}{4} when ± is plus. Add 8 to 14\sqrt{2}.

$$x=\frac{14\sqrt{2}+8}{4}$$

Divide 8+14\sqrt{2} by 4.

$$x=\frac{7\sqrt{2}}{2}+2$$

Now solve the equation x=\frac{8±14\sqrt{2}}{4} when ± is minus. Subtract 14\sqrt{2} from 8.

$$x=\frac{8-14\sqrt{2}}{4}$$

Divide 8-14\sqrt{2} by 4.

$$x=-\frac{7\sqrt{2}}{2}+2$$

The equation is now solved.

$$x=\frac{7\sqrt{2}}{2}+2 x=-\frac{7\sqrt{2}}{2}+2$$

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}.

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

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

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

Combine x^{2} and x^{2} to get 2x^{2}.

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

Combine -4x and -4x to get -8x.

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

Add 4 and 4 to get 8.

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

Calculate 7 to the power of 2 and get 49.

$$2x^{2}-8x+8=49$$

Subtract 8 from both sides.

$$2x^{2}-8x=49-8$$

Subtract 8 from 49 to get 41.

$$2x^{2}-8x=41$$

Divide both sides by 2.

$$\frac{2x^{2}-8x}{2}=\frac{41}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{8}{2}\right)x=\frac{41}{2}$$

Divide -8 by 2.

$$x^{2}-4x=\frac{41}{2}$$

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.

$$x^{2}-4x+\left(-2\right)^{2}=\frac{41}{2}+\left(-2\right)^{2}$$

Square -2.

$$x^{2}-4x+4=\frac{41}{2}+4$$

Add \frac{41}{2} to 4.

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

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}.

$$\left(x-2\right)^{2}=\frac{49}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{\frac{49}{2}}$$

Simplify.

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

Add 2 to both sides of the equation.

$$x=\frac{7\sqrt{2}}{2}+2 x=-\frac{7\sqrt{2}}{2}+2$$