Solve 0=(x-5)^2-6

$$0 = ( x - 5 ) ^ { 2 } - 6$$

Solve for x

$$x=\sqrt{6}+5\approx 7.449489743$$

$$x=5-\sqrt{6}\approx 2.550510257$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$0=x^{2}-10x+25-6$$

Subtract 6 from 25 to get 19.

$$0=x^{2}-10x+19$$

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

$$x^{2}-10x+19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and 19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 19}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 19}}{2}$$

Multiply -4 times 19.

$$x=\frac{-\left(-10\right)±\sqrt{100-76}}{2}$$

Add 100 to -76.

$$x=\frac{-\left(-10\right)±\sqrt{24}}{2}$$

Take the square root of 24.

$$x=\frac{-\left(-10\right)±2\sqrt{6}}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±2\sqrt{6}}{2}$$

Now solve the equation x=\frac{10±2\sqrt{6}}{2} when ± is plus. Add 10 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}+10}{2}$$

Divide 10+2\sqrt{6} by 2.

$$x=\sqrt{6}+5$$

Now solve the equation x=\frac{10±2\sqrt{6}}{2} when ± is minus. Subtract 2\sqrt{6} from 10.

$$x=\frac{10-2\sqrt{6}}{2}$$

Divide 10-2\sqrt{6} by 2.

$$x=5-\sqrt{6}$$

The equation is now solved.

$$x=\sqrt{6}+5 x=5-\sqrt{6}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$0=x^{2}-10x+25-6$$

Subtract 6 from 25 to get 19.

$$0=x^{2}-10x+19$$

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

$$x^{2}-10x+19=0$$

Subtract 19 from both sides. Anything subtracted from zero gives its negation.

$$x^{2}-10x=-19$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-19+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-19+25$$

Add -19 to 25.

$$x^{2}-10x+25=6$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=6$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-5=\sqrt{6} x-5=-\sqrt{6}$$

Add 5 to both sides of the equation.

$$x=\sqrt{6}+5 x=5-\sqrt{6}$$