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

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

Solve for x

$$x=28$$

$$x=0$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+7\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$2x^{2}-14x+49=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=14x+49$$

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

$$x^{2}-14x+49=14x+49$$

Subtract 14x from both sides.

$$x^{2}-14x+49-14x=49$$

Combine -14x and -14x to get -28x.

$$x^{2}-28x+49=49$$

Subtract 49 from both sides.

$$x^{2}-28x+49-49=0$$

Subtract 49 from 49 to get 0.

$$x^{2}-28x=0$$

Factor out x.

$$x\left(x-28\right)=0$$

To find equation solutions, solve x=0 and x-28=0.

$$x=0 x=28$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+7\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$2x^{2}-14x+49=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=14x+49$$

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

$$x^{2}-14x+49=14x+49$$

Subtract 14x from both sides.

$$x^{2}-14x+49-14x=49$$

Combine -14x and -14x to get -28x.

$$x^{2}-28x+49=49$$

Subtract 49 from both sides.

$$x^{2}-28x+49-49=0$$

Subtract 49 from 49 to get 0.

$$x^{2}-28x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -28 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-28\right)±\sqrt{\left(-28\right)^{2}}}{2}$$

Take the square root of \left(-28\right)^{2}.

$$x=\frac{-\left(-28\right)±28}{2}$$

The opposite of -28 is 28.

$$x=\frac{28±28}{2}$$

Now solve the equation x=\frac{28±28}{2} when ± is plus. Add 28 to 28.

$$x=\frac{56}{2}$$

Divide 56 by 2.

$$x=28$$

Now solve the equation x=\frac{28±28}{2} when ± is minus. Subtract 28 from 28.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=28 x=0$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-7\right)^{2}.

$$x^{2}-14x+49+x^{2}=\left(x+7\right)^{2}$$

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

$$2x^{2}-14x+49=\left(x+7\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+7\right)^{2}.

$$2x^{2}-14x+49=x^{2}+14x+49$$

Subtract x^{2} from both sides.

$$2x^{2}-14x+49-x^{2}=14x+49$$

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

$$x^{2}-14x+49=14x+49$$

Subtract 14x from both sides.

$$x^{2}-14x+49-14x=49$$

Combine -14x and -14x to get -28x.

$$x^{2}-28x+49=49$$

Subtract 49 from both sides.

$$x^{2}-28x+49-49=0$$

Subtract 49 from 49 to get 0.

$$x^{2}-28x=0$$

Divide -28, the coefficient of the x term, by 2 to get -14. Then add the square of -14 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-28x+\left(-14\right)^{2}=\left(-14\right)^{2}$$

Square -14.

$$x^{2}-28x+196=196$$

Factor x^{2}-28x+196. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-14\right)^{2}=196$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-14\right)^{2}}=\sqrt{196}$$

Simplify.

$$x-14=14 x-14=-14$$

Add 14 to both sides of the equation.

$$x=28 x=0$$