Solve 6^2+(13-x)^2=(2x)^2

$$6 ^ { 2 } + ( 13 - x ) ^ { 2 } = ( 2 x ) ^ { 2 }$$

Solve for x

$$x = -\frac{41}{3} = -13\frac{2}{3} \approx -13.666666667$$

$$x=5$$

Steps Using Factoring By Grouping

Calculate 6 to the power of 2 and get 36.

$$36+\left(13-x\right)^{2}=\left(2x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(13-x\right)^{2}.

$$36+169-26x+x^{2}=\left(2x\right)^{2}$$

Add 36 and 169 to get 205.

$$205-26x+x^{2}=\left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$205-26x+x^{2}=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$205-26x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$205-26x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$205-26x-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}-26x+205=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+205. To find a and b, set up a system to be solved.

$$a+b=-26 ab=-3\times 205=-615$$

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. List all such integer pairs that give product -615.

$$1,-615 3,-205 5,-123 15,-41$$

Calculate the sum for each pair.

$$1-615=-614 3-205=-202 5-123=-118 15-41=-26$$

The solution is the pair that gives sum -26.

$$a=15 b=-41$$

Rewrite -3x^{2}-26x+205 as \left(-3x^{2}+15x\right)+\left(-41x+205\right).

$$\left(-3x^{2}+15x\right)+\left(-41x+205\right)$$

Factor out 3x in the first and 41 in the second group.

$$3x\left(-x+5\right)+41\left(-x+5\right)$$

Factor out common term -x+5 by using distributive property.

$$\left(-x+5\right)\left(3x+41\right)$$

To find equation solutions, solve -x+5=0 and 3x+41=0.

$$x=5 x=-\frac{41}{3}$$

Steps Using the Quadratic Formula

Calculate 6 to the power of 2 and get 36.

$$36+\left(13-x\right)^{2}=\left(2x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(13-x\right)^{2}.

$$36+169-26x+x^{2}=\left(2x\right)^{2}$$

Add 36 and 169 to get 205.

$$205-26x+x^{2}=\left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$205-26x+x^{2}=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$205-26x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$205-26x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$205-26x-3x^{2}=0$$

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.

$$-3x^{2}-26x+205=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -26 for b, and 205 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-26\right)±\sqrt{\left(-26\right)^{2}-4\left(-3\right)\times 205}}{2\left(-3\right)}$$

Square -26.

$$x=\frac{-\left(-26\right)±\sqrt{676-4\left(-3\right)\times 205}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-26\right)±\sqrt{676+12\times 205}}{2\left(-3\right)}$$

Multiply 12 times 205.

$$x=\frac{-\left(-26\right)±\sqrt{676+2460}}{2\left(-3\right)}$$

Add 676 to 2460.

$$x=\frac{-\left(-26\right)±\sqrt{3136}}{2\left(-3\right)}$$

Take the square root of 3136.

$$x=\frac{-\left(-26\right)±56}{2\left(-3\right)}$$

The opposite of -26 is 26.

$$x=\frac{26±56}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{26±56}{-6}$$

Now solve the equation x=\frac{26±56}{-6} when ± is plus. Add 26 to 56.

$$x=\frac{82}{-6}$$

Reduce the fraction \frac{82}{-6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{41}{3}$$

Now solve the equation x=\frac{26±56}{-6} when ± is minus. Subtract 56 from 26.

$$x=-\frac{30}{-6}$$

Divide -30 by -6.

$$x=5$$

The equation is now solved.

$$x=-\frac{41}{3} x=5$$

Steps for Completing the Square

Calculate 6 to the power of 2 and get 36.

$$36+\left(13-x\right)^{2}=\left(2x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(13-x\right)^{2}.

$$36+169-26x+x^{2}=\left(2x\right)^{2}$$

Add 36 and 169 to get 205.

$$205-26x+x^{2}=\left(2x\right)^{2}$$

Expand \left(2x\right)^{2}.

$$205-26x+x^{2}=2^{2}x^{2}$$

Calculate 2 to the power of 2 and get 4.

$$205-26x+x^{2}=4x^{2}$$

Subtract 4x^{2} from both sides.

$$205-26x+x^{2}-4x^{2}=0$$

Combine x^{2} and -4x^{2} to get -3x^{2}.

$$205-26x-3x^{2}=0$$

Subtract 205 from both sides. Anything subtracted from zero gives its negation.

$$-26x-3x^{2}=-205$$

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.

$$-3x^{2}-26x=-205$$

Divide both sides by -3.

$$\frac{-3x^{2}-26x}{-3}=-\frac{205}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{26}{-3}\right)x=-\frac{205}{-3}$$

Divide -26 by -3.

$$x^{2}+\frac{26}{3}x=-\frac{205}{-3}$$

Divide -205 by -3.

$$x^{2}+\frac{26}{3}x=\frac{205}{3}$$

Divide \frac{26}{3}, the coefficient of the x term, by 2 to get \frac{13}{3}. Then add the square of \frac{13}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{26}{3}x+\left(\frac{13}{3}\right)^{2}=\frac{205}{3}+\left(\frac{13}{3}\right)^{2}$$

Square \frac{13}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{26}{3}x+\frac{169}{9}=\frac{205}{3}+\frac{169}{9}$$

Add \frac{205}{3} to \frac{169}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{26}{3}x+\frac{169}{9}=\frac{784}{9}$$

Factor x^{2}+\frac{26}{3}x+\frac{169}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{3}\right)^{2}=\frac{784}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{3}\right)^{2}}=\sqrt{\frac{784}{9}}$$

Simplify.

$$x+\frac{13}{3}=\frac{28}{3} x+\frac{13}{3}=-\frac{28}{3}$$

Subtract \frac{13}{3} from both sides of the equation.

$$x=5 x=-\frac{41}{3}$$