Solve (x-4)^2-4(4+0,5)=2

$$( x - 4 ) ^ { 2 } - 4 ( 4 + 0,5 ) = 2$$

Solve for x

$$x=2\sqrt{5}+4\approx 8.472135955$$

$$x=4-2\sqrt{5}\approx -0.472135955$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-4\left(4+0,5\right)=2$$

Add 4 and 0,5 to get 4,5.

$$x^{2}-8x+16-4\times 4,5=2$$

Multiply 4 and 4,5 to get 18.

$$x^{2}-8x+16-18=2$$

Subtract 18 from 16 to get -2.

$$x^{2}-8x-2=2$$

Subtract 2 from both sides.

$$x^{2}-8x-2-2=0$$

Subtract 2 from -2 to get -4.

$$x^{2}-8x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -8 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-8\right)±\sqrt{\left(-8\right)^{2}-4\left(-4\right)}}{2}$$

Square -8.

$$x=\frac{-\left(-8\right)±\sqrt{64-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$x=\frac{-\left(-8\right)±\sqrt{64+16}}{2}$$

Add 64 to 16.

$$x=\frac{-\left(-8\right)±\sqrt{80}}{2}$$

Take the square root of 80.

$$x=\frac{-\left(-8\right)±4\sqrt{5}}{2}$$

The opposite of -8 is 8.

$$x=\frac{8±4\sqrt{5}}{2}$$

Now solve the equation x=\frac{8±4\sqrt{5}}{2} when ± is plus. Add 8 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}+8}{2}$$

Divide 8+4\sqrt{5} by 2.

$$x=2\sqrt{5}+4$$

Now solve the equation x=\frac{8±4\sqrt{5}}{2} when ± is minus. Subtract 4\sqrt{5} from 8.

$$x=\frac{8-4\sqrt{5}}{2}$$

Divide 8-4\sqrt{5} by 2.

$$x=4-2\sqrt{5}$$

The equation is now solved.

$$x=2\sqrt{5}+4 x=4-2\sqrt{5}$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-4\right)^{2}.

$$x^{2}-8x+16-4\left(4+0,5\right)=2$$

Add 4 and 0,5 to get 4,5.

$$x^{2}-8x+16-4\times 4,5=2$$

Multiply 4 and 4,5 to get 18.

$$x^{2}-8x+16-18=2$$

Subtract 18 from 16 to get -2.

$$x^{2}-8x-2=2$$

Add 2 to both sides.

$$x^{2}-8x=2+2$$

Add 2 and 2 to get 4.

$$x^{2}-8x=4$$

Divide -8, the coefficient of the x term, by 2 to get -4. Then add the square of -4 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-8x+\left(-4\right)^{2}=4+\left(-4\right)^{2}$$

Square -4.

$$x^{2}-8x+16=4+16$$

Add 4 to 16.

$$x^{2}-8x+16=20$$

Factor x^{2}-8x+16. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-4\right)^{2}=20$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-4\right)^{2}}=\sqrt{20}$$

Simplify.

$$x-4=2\sqrt{5} x-4=-2\sqrt{5}$$

Add 4 to both sides of the equation.

$$x=2\sqrt{5}+4 x=4-2\sqrt{5}$$