Solve 26^2=x^2+(x+14)^2

$$26 ^ { 2 } = x ^ { 2 } + ( x + 14 ) ^ { 2 }$$

Solve for x

$$x=-24$$

$$x=10$$

Steps Using Factoring By Grouping

Calculate 26 to the power of 2 and get 676.

$$676=x^{2}+\left(x+14\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+14\right)^{2}.

$$676=x^{2}+x^{2}+28x+196$$

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

$$676=2x^{2}+28x+196$$

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

$$2x^{2}+28x+196=676$$

Subtract 676 from both sides.

$$2x^{2}+28x+196-676=0$$

Subtract 676 from 196 to get -480.

$$2x^{2}+28x-480=0$$

Divide both sides by 2.

$$x^{2}+14x-240=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-240. To find a and b, set up a system to be solved.

$$a+b=14 ab=1\left(-240\right)=-240$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -240.

$$-1,240 -2,120 -3,80 -4,60 -5,48 -6,40 -8,30 -10,24 -12,20 -15,16$$

Calculate the sum for each pair.

$$-1+240=239 -2+120=118 -3+80=77 -4+60=56 -5+48=43 -6+40=34 -8+30=22 -10+24=14 -12+20=8 -15+16=1$$

The solution is the pair that gives sum 14.

$$a=-10 b=24$$

Rewrite x^{2}+14x-240 as \left(x^{2}-10x\right)+\left(24x-240\right).

$$\left(x^{2}-10x\right)+\left(24x-240\right)$$

Factor out x in the first and 24 in the second group.

$$x\left(x-10\right)+24\left(x-10\right)$$

Factor out common term x-10 by using distributive property.

$$\left(x-10\right)\left(x+24\right)$$

To find equation solutions, solve x-10=0 and x+24=0.

$$x=10 x=-24$$

Steps Using the Quadratic Formula

Calculate 26 to the power of 2 and get 676.

$$676=x^{2}+\left(x+14\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+14\right)^{2}.

$$676=x^{2}+x^{2}+28x+196$$

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

$$676=2x^{2}+28x+196$$

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

$$2x^{2}+28x+196=676$$

Subtract 676 from both sides.

$$2x^{2}+28x+196-676=0$$

Subtract 676 from 196 to get -480.

$$2x^{2}+28x-480=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 28 for b, and -480 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-28±\sqrt{28^{2}-4\times 2\left(-480\right)}}{2\times 2}$$

Square 28.

$$x=\frac{-28±\sqrt{784-4\times 2\left(-480\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-28±\sqrt{784-8\left(-480\right)}}{2\times 2}$$

Multiply -8 times -480.

$$x=\frac{-28±\sqrt{784+3840}}{2\times 2}$$

Add 784 to 3840.

$$x=\frac{-28±\sqrt{4624}}{2\times 2}$$

Take the square root of 4624.

$$x=\frac{-28±68}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-28±68}{4}$$

Now solve the equation x=\frac{-28±68}{4} when ± is plus. Add -28 to 68.

$$x=\frac{40}{4}$$

Divide 40 by 4.

$$x=10$$

Now solve the equation x=\frac{-28±68}{4} when ± is minus. Subtract 68 from -28.

$$x=-\frac{96}{4}$$

Divide -96 by 4.

$$x=-24$$

The equation is now solved.

$$x=10 x=-24$$

Steps for Completing the Square

Calculate 26 to the power of 2 and get 676.

$$676=x^{2}+\left(x+14\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+14\right)^{2}.

$$676=x^{2}+x^{2}+28x+196$$

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

$$676=2x^{2}+28x+196$$

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

$$2x^{2}+28x+196=676$$

Subtract 196 from both sides.

$$2x^{2}+28x=676-196$$

Subtract 196 from 676 to get 480.

$$2x^{2}+28x=480$$

Divide both sides by 2.

$$\frac{2x^{2}+28x}{2}=\frac{480}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{28}{2}x=\frac{480}{2}$$

Divide 28 by 2.

$$x^{2}+14x=\frac{480}{2}$$

Divide 480 by 2.

$$x^{2}+14x=240$$

Divide 14, the coefficient of the x term, by 2 to get 7. Then add the square of 7 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+14x+7^{2}=240+7^{2}$$

Square 7.

$$x^{2}+14x+49=240+49$$

Add 240 to 49.

$$x^{2}+14x+49=289$$

Factor x^{2}+14x+49. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+7\right)^{2}=289$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+7\right)^{2}}=\sqrt{289}$$

Simplify.

$$x+7=17 x+7=-17$$

Subtract 7 from both sides of the equation.

$$x=10 x=-24$$