Solve left(x+14right)^2-left(x+11right)^2=left(x-6right)^2

$${ \left(x+14 \right) }^{ 2 } - { \left(x+11 \right) }^{ 2 } = { \left(x-6 \right) }^{ 2 }$$

Solve for x

$$x=2\sqrt{30}+9\approx 19.95445115$$

$$x=9-2\sqrt{30}\approx -1.95445115$$

Steps Using the Quadratic Formula

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

$$x^{2}+28x+196-\left(x+11\right)^{2}=\left(x-6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+11\right)^{2}.

$$x^{2}+28x+196-\left(x^{2}+22x+121\right)=\left(x-6\right)^{2}$$

To find the opposite of x^{2}+22x+121, find the opposite of each term.

$$x^{2}+28x+196-x^{2}-22x-121=\left(x-6\right)^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$28x+196-22x-121=\left(x-6\right)^{2}$$

Combine 28x and -22x to get 6x.

$$6x+196-121=\left(x-6\right)^{2}$$

Subtract 121 from 196 to get 75.

$$6x+75=\left(x-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$6x+75=x^{2}-12x+36$$

Subtract x^{2} from both sides.

$$6x+75-x^{2}=-12x+36$$

Add 12x to both sides.

$$6x+75-x^{2}+12x=36$$

Combine 6x and 12x to get 18x.

$$18x+75-x^{2}=36$$

Subtract 36 from both sides.

$$18x+75-x^{2}-36=0$$

Subtract 36 from 75 to get 39.

$$18x+39-x^{2}=0$$

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.

$$-x^{2}+18x+39=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 18 for b, and 39 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-18±\sqrt{18^{2}-4\left(-1\right)\times 39}}{2\left(-1\right)}$$

Square 18.

$$x=\frac{-18±\sqrt{324-4\left(-1\right)\times 39}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-18±\sqrt{324+4\times 39}}{2\left(-1\right)}$$

Multiply 4 times 39.

$$x=\frac{-18±\sqrt{324+156}}{2\left(-1\right)}$$

Add 324 to 156.

$$x=\frac{-18±\sqrt{480}}{2\left(-1\right)}$$

Take the square root of 480.

$$x=\frac{-18±4\sqrt{30}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-18±4\sqrt{30}}{-2}$$

Now solve the equation x=\frac{-18±4\sqrt{30}}{-2} when ± is plus. Add -18 to 4\sqrt{30}.

$$x=\frac{4\sqrt{30}-18}{-2}$$

Divide -18+4\sqrt{30} by -2.

$$x=9-2\sqrt{30}$$

Now solve the equation x=\frac{-18±4\sqrt{30}}{-2} when ± is minus. Subtract 4\sqrt{30} from -18.

$$x=\frac{-4\sqrt{30}-18}{-2}$$

Divide -18-4\sqrt{30} by -2.

$$x=2\sqrt{30}+9$$

The equation is now solved.

$$x=9-2\sqrt{30} x=2\sqrt{30}+9$$

Steps for Completing the Square

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

$$x^{2}+28x+196-\left(x+11\right)^{2}=\left(x-6\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(x+11\right)^{2}.

$$x^{2}+28x+196-\left(x^{2}+22x+121\right)=\left(x-6\right)^{2}$$

To find the opposite of x^{2}+22x+121, find the opposite of each term.

$$x^{2}+28x+196-x^{2}-22x-121=\left(x-6\right)^{2}$$

Combine x^{2} and -x^{2} to get 0.

$$28x+196-22x-121=\left(x-6\right)^{2}$$

Combine 28x and -22x to get 6x.

$$6x+196-121=\left(x-6\right)^{2}$$

Subtract 121 from 196 to get 75.

$$6x+75=\left(x-6\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-6\right)^{2}.

$$6x+75=x^{2}-12x+36$$

Subtract x^{2} from both sides.

$$6x+75-x^{2}=-12x+36$$

Add 12x to both sides.

$$6x+75-x^{2}+12x=36$$

Combine 6x and 12x to get 18x.

$$18x+75-x^{2}=36$$

Subtract 75 from both sides.

$$18x-x^{2}=36-75$$

Subtract 75 from 36 to get -39.

$$18x-x^{2}=-39$$

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.

$$-x^{2}+18x=-39$$

Divide both sides by -1.

$$\frac{-x^{2}+18x}{-1}=-\frac{39}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{18}{-1}x=-\frac{39}{-1}$$

Divide 18 by -1.

$$x^{2}-18x=-\frac{39}{-1}$$

Divide -39 by -1.

$$x^{2}-18x=39$$

Divide -18, the coefficient of the x term, by 2 to get -9. Then add the square of -9 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-18x+\left(-9\right)^{2}=39+\left(-9\right)^{2}$$

Square -9.

$$x^{2}-18x+81=39+81$$

Add 39 to 81.

$$x^{2}-18x+81=120$$

Factor x^{2}-18x+81. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-9\right)^{2}=120$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-9\right)^{2}}=\sqrt{120}$$

Simplify.

$$x-9=2\sqrt{30} x-9=-2\sqrt{30}$$

Add 9 to both sides of the equation.

$$x=2\sqrt{30}+9 x=9-2\sqrt{30}$$