Solve left(7+xright)^2+x^2=left(18-xright)^2

$${ \left(7+x \right) }^{ 2 } + { x }^{ 2 } = { \left(18-x \right) }^{ 2 }$$

Solve for x

$$x=-55$$

$$x=5$$

Steps Using Factoring

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

$$49+14x+x^{2}+x^{2}=\left(18-x\right)^{2}$$

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

$$49+14x+2x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$49+14x+2x^{2}=324-36x+x^{2}$$

Subtract 324 from both sides.

$$49+14x+2x^{2}-324=-36x+x^{2}$$

Subtract 324 from 49 to get -275.

$$-275+14x+2x^{2}=-36x+x^{2}$$

Add 36x to both sides.

$$-275+14x+2x^{2}+36x=x^{2}$$

Combine 14x and 36x to get 50x.

$$-275+50x+2x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$-275+50x+2x^{2}-x^{2}=0$$

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

$$-275+50x+x^{2}=0$$

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

$$x^{2}+50x-275=0$$

To solve the equation, factor x^{2}+50x-275 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=50 ab=-275$$

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 -275.

$$-1,275 -5,55 -11,25$$

Calculate the sum for each pair.

$$-1+275=274 -5+55=50 -11+25=14$$

The solution is the pair that gives sum 50.

$$a=-5 b=55$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-5\right)\left(x+55\right)$$

To find equation solutions, solve x-5=0 and x+55=0.

$$x=5 x=-55$$

Steps Using Factoring By Grouping

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

$$49+14x+x^{2}+x^{2}=\left(18-x\right)^{2}$$

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

$$49+14x+2x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$49+14x+2x^{2}=324-36x+x^{2}$$

Subtract 324 from both sides.

$$49+14x+2x^{2}-324=-36x+x^{2}$$

Subtract 324 from 49 to get -275.

$$-275+14x+2x^{2}=-36x+x^{2}$$

Add 36x to both sides.

$$-275+14x+2x^{2}+36x=x^{2}$$

Combine 14x and 36x to get 50x.

$$-275+50x+2x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$-275+50x+2x^{2}-x^{2}=0$$

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

$$-275+50x+x^{2}=0$$

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

$$x^{2}+50x-275=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-275. To find a and b, set up a system to be solved.

$$a+b=50 ab=1\left(-275\right)=-275$$

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 -275.

$$-1,275 -5,55 -11,25$$

Calculate the sum for each pair.

$$-1+275=274 -5+55=50 -11+25=14$$

The solution is the pair that gives sum 50.

$$a=-5 b=55$$

Rewrite x^{2}+50x-275 as \left(x^{2}-5x\right)+\left(55x-275\right).

$$\left(x^{2}-5x\right)+\left(55x-275\right)$$

Factor out x in the first and 55 in the second group.

$$x\left(x-5\right)+55\left(x-5\right)$$

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

$$\left(x-5\right)\left(x+55\right)$$

To find equation solutions, solve x-5=0 and x+55=0.

$$x=5 x=-55$$

Steps Using the Quadratic Formula

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

$$49+14x+x^{2}+x^{2}=\left(18-x\right)^{2}$$

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

$$49+14x+2x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$49+14x+2x^{2}=324-36x+x^{2}$$

Subtract 324 from both sides.

$$49+14x+2x^{2}-324=-36x+x^{2}$$

Subtract 324 from 49 to get -275.

$$-275+14x+2x^{2}=-36x+x^{2}$$

Add 36x to both sides.

$$-275+14x+2x^{2}+36x=x^{2}$$

Combine 14x and 36x to get 50x.

$$-275+50x+2x^{2}=x^{2}$$

Subtract x^{2} from both sides.

$$-275+50x+2x^{2}-x^{2}=0$$

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

$$-275+50x+x^{2}=0$$

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.

$$x^{2}+50x-275=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 50 for b, and -275 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-50±\sqrt{50^{2}-4\left(-275\right)}}{2}$$

Square 50.

$$x=\frac{-50±\sqrt{2500-4\left(-275\right)}}{2}$$

Multiply -4 times -275.

$$x=\frac{-50±\sqrt{2500+1100}}{2}$$

Add 2500 to 1100.

$$x=\frac{-50±\sqrt{3600}}{2}$$

Take the square root of 3600.

$$x=\frac{-50±60}{2}$$

Now solve the equation x=\frac{-50±60}{2} when ± is plus. Add -50 to 60.

$$x=\frac{10}{2}$$

Divide 10 by 2.

$$x=5$$

Now solve the equation x=\frac{-50±60}{2} when ± is minus. Subtract 60 from -50.

$$x=-\frac{110}{2}$$

Divide -110 by 2.

$$x=-55$$

The equation is now solved.

$$x=5 x=-55$$

Steps for Completing the Square

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

$$49+14x+x^{2}+x^{2}=\left(18-x\right)^{2}$$

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

$$49+14x+2x^{2}=\left(18-x\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(18-x\right)^{2}.

$$49+14x+2x^{2}=324-36x+x^{2}$$

Add 36x to both sides.

$$49+14x+2x^{2}+36x=324+x^{2}$$

Combine 14x and 36x to get 50x.

$$49+50x+2x^{2}=324+x^{2}$$

Subtract x^{2} from both sides.

$$49+50x+2x^{2}-x^{2}=324$$

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

$$49+50x+x^{2}=324$$

Subtract 49 from both sides.

$$50x+x^{2}=324-49$$

Subtract 49 from 324 to get 275.

$$50x+x^{2}=275$$

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.

$$x^{2}+50x=275$$

Divide 50, the coefficient of the x term, by 2 to get 25. Then add the square of 25 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+50x+25^{2}=275+25^{2}$$

Square 25.

$$x^{2}+50x+625=275+625$$

Add 275 to 625.

$$x^{2}+50x+625=900$$

Factor x^{2}+50x+625. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+25\right)^{2}=900$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+25\right)^{2}}=\sqrt{900}$$

Simplify.

$$x+25=30 x+25=-30$$

Subtract 25 from both sides of the equation.

$$x=5 x=-55$$