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

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

Solve for x

$$x=-\frac{\sqrt{46}}{2}+6\approx 2.608835008$$

$$x=\frac{\sqrt{46}}{2}+6\approx 9.391164992$$

Steps Using the Quadratic Formula

Calculate 5 to the power of 2 and get 25.

$$25=\left(7-x\right)^{2}+\left(5-x\right)^{2}$$

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

$$25=49-14x+x^{2}+\left(5-x\right)^{2}$$

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

$$25=49-14x+x^{2}+25-10x+x^{2}$$

Add 49 and 25 to get 74.

$$25=74-14x+x^{2}-10x+x^{2}$$

Combine -14x and -10x to get -24x.

$$25=74-24x+x^{2}+x^{2}$$

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

$$25=74-24x+2x^{2}$$

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

$$74-24x+2x^{2}=25$$

Subtract 25 from both sides.

$$74-24x+2x^{2}-25=0$$

Subtract 25 from 74 to get 49.

$$49-24x+2x^{2}=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$2x^{2}-24x+49=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -24 for b, and 49 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-24\right)±\sqrt{\left(-24\right)^{2}-4\times 2\times 49}}{2\times 2}$$

Square -24.

$$x=\frac{-\left(-24\right)±\sqrt{576-4\times 2\times 49}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-24\right)±\sqrt{576-8\times 49}}{2\times 2}$$

Multiply -8 times 49.

$$x=\frac{-\left(-24\right)±\sqrt{576-392}}{2\times 2}$$

Add 576 to -392.

$$x=\frac{-\left(-24\right)±\sqrt{184}}{2\times 2}$$

Take the square root of 184.

$$x=\frac{-\left(-24\right)±2\sqrt{46}}{2\times 2}$$

The opposite of -24 is 24.

$$x=\frac{24±2\sqrt{46}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{24±2\sqrt{46}}{4}$$

Now solve the equation x=\frac{24±2\sqrt{46}}{4} when ± is plus. Add 24 to 2\sqrt{46}.

$$x=\frac{2\sqrt{46}+24}{4}$$

Divide 24+2\sqrt{46} by 4.

$$x=\frac{\sqrt{46}}{2}+6$$

Now solve the equation x=\frac{24±2\sqrt{46}}{4} when ± is minus. Subtract 2\sqrt{46} from 24.

$$x=\frac{24-2\sqrt{46}}{4}$$

Divide 24-2\sqrt{46} by 4.

$$x=-\frac{\sqrt{46}}{2}+6$$

The equation is now solved.

$$x=\frac{\sqrt{46}}{2}+6 x=-\frac{\sqrt{46}}{2}+6$$

Steps for Completing the Square

Calculate 5 to the power of 2 and get 25.

$$25=\left(7-x\right)^{2}+\left(5-x\right)^{2}$$

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

$$25=49-14x+x^{2}+\left(5-x\right)^{2}$$

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

$$25=49-14x+x^{2}+25-10x+x^{2}$$

Add 49 and 25 to get 74.

$$25=74-14x+x^{2}-10x+x^{2}$$

Combine -14x and -10x to get -24x.

$$25=74-24x+x^{2}+x^{2}$$

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

$$25=74-24x+2x^{2}$$

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

$$74-24x+2x^{2}=25$$

Subtract 74 from both sides.

$$-24x+2x^{2}=25-74$$

Subtract 74 from 25 to get -49.

$$-24x+2x^{2}=-49$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-24x=-49$$

Divide both sides by 2.

$$\frac{2x^{2}-24x}{2}=-\frac{49}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\left(-\frac{24}{2}\right)x=-\frac{49}{2}$$

Divide -24 by 2.

$$x^{2}-12x=-\frac{49}{2}$$

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.

$$x^{2}-12x+\left(-6\right)^{2}=-\frac{49}{2}+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-\frac{49}{2}+36$$

Add -\frac{49}{2} to 36.

$$x^{2}-12x+36=\frac{23}{2}$$

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}.

$$\left(x-6\right)^{2}=\frac{23}{2}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{\frac{23}{2}}$$

Simplify.

$$x-6=\frac{\sqrt{46}}{2} x-6=-\frac{\sqrt{46}}{2}$$

Add 6 to both sides of the equation.

$$x=\frac{\sqrt{46}}{2}+6 x=-\frac{\sqrt{46}}{2}+6$$