Solve (x-11)^2-5-11^2=0

$$( x - 11 ) ^ { 2 } - 5 - 11 ^ { 2 } = 0$$

Solve for x

$$x=3\sqrt{14}+11\approx 22.22497216$$

$$x=11-3\sqrt{14}\approx -0.22497216$$

Steps Using the Quadratic Formula

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

$$x^{2}-22x+121-5-11^{2}=0$$

Subtract 5 from 121 to get 116.

$$x^{2}-22x+116-11^{2}=0$$

Calculate 11 to the power of 2 and get 121.

$$x^{2}-22x+116-121=0$$

Subtract 121 from 116 to get -5.

$$x^{2}-22x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -22 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-22\right)±\sqrt{\left(-22\right)^{2}-4\left(-5\right)}}{2}$$

Square -22.

$$x=\frac{-\left(-22\right)±\sqrt{484-4\left(-5\right)}}{2}$$

Multiply -4 times -5.

$$x=\frac{-\left(-22\right)±\sqrt{484+20}}{2}$$

Add 484 to 20.

$$x=\frac{-\left(-22\right)±\sqrt{504}}{2}$$

Take the square root of 504.

$$x=\frac{-\left(-22\right)±6\sqrt{14}}{2}$$

The opposite of -22 is 22.

$$x=\frac{22±6\sqrt{14}}{2}$$

Now solve the equation x=\frac{22±6\sqrt{14}}{2} when ± is plus. Add 22 to 6\sqrt{14}.

$$x=\frac{6\sqrt{14}+22}{2}$$

Divide 22+6\sqrt{14} by 2.

$$x=3\sqrt{14}+11$$

Now solve the equation x=\frac{22±6\sqrt{14}}{2} when ± is minus. Subtract 6\sqrt{14} from 22.

$$x=\frac{22-6\sqrt{14}}{2}$$

Divide 22-6\sqrt{14} by 2.

$$x=11-3\sqrt{14}$$

The equation is now solved.

$$x=3\sqrt{14}+11 x=11-3\sqrt{14}$$

Steps for Completing the Square

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

$$x^{2}-22x+121-5-11^{2}=0$$

Subtract 5 from 121 to get 116.

$$x^{2}-22x+116-11^{2}=0$$

Calculate 11 to the power of 2 and get 121.

$$x^{2}-22x+116-121=0$$

Subtract 121 from 116 to get -5.

$$x^{2}-22x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$x^{2}-22x=5$$

Divide -22, the coefficient of the x term, by 2 to get -11. Then add the square of -11 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-22x+\left(-11\right)^{2}=5+\left(-11\right)^{2}$$

Square -11.

$$x^{2}-22x+121=5+121$$

Add 5 to 121.

$$x^{2}-22x+121=126$$

Factor x^{2}-22x+121. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-11\right)^{2}=126$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-11\right)^{2}}=\sqrt{126}$$

Simplify.

$$x-11=3\sqrt{14} x-11=-3\sqrt{14}$$

Add 11 to both sides of the equation.

$$x=3\sqrt{14}+11 x=11-3\sqrt{14}$$