Solve 2x^2+(8-sqrt{2})=4sqrt{2}

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

Solve for x (complex solution)

$$x=-\frac{i\sqrt{16-10\sqrt{2}}}{2}\approx -0-0.681517494i$$

$$x=\frac{i\sqrt{16-10\sqrt{2}}}{2}\approx 0.681517494i$$

Steps by Finding Square Root

Subtract 8 from both sides.

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

Add \sqrt{2} to both sides.

$$2x^{2}=4\sqrt{2}-8+\sqrt{2}$$

Combine 4\sqrt{2} and \sqrt{2} to get 5\sqrt{2}.

$$2x^{2}=5\sqrt{2}-8$$

Dividing by 2 undoes the multiplication by 2.

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

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

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

Take the square root of both sides of the equation.

$$x=\frac{i\sqrt{16-10\sqrt{2}}}{2} x=-\frac{i\sqrt{16-10\sqrt{2}}}{2}$$

Steps Using the Quadratic Formula

Subtract 4\sqrt{2} from both sides.

$$2x^{2}+8-\sqrt{2}-4\sqrt{2}=0$$

Combine -\sqrt{2} and -4\sqrt{2} to get -5\sqrt{2}.

$$2x^{2}+8-5\sqrt{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 0 for b, and 8-5\sqrt{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{0±\sqrt{0^{2}-4\times 2\left(8-5\sqrt{2}\right)}}{2\times 2}$$

Square 0.

$$x=\frac{0±\sqrt{-4\times 2\left(8-5\sqrt{2}\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{0±\sqrt{-8\left(8-5\sqrt{2}\right)}}{2\times 2}$$

Multiply -8 times 8-5\sqrt{2}.

$$x=\frac{0±\sqrt{40\sqrt{2}-64}}{2\times 2}$$

Take the square root of -64+40\sqrt{2}.

$$x=\frac{0±2i\sqrt{16-10\sqrt{2}}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{0±2i\sqrt{16-10\sqrt{2}}}{4}$$

Now solve the equation x=\frac{0±2i\sqrt{16-10\sqrt{2}}}{4} when ± is plus.

$$x=\frac{i\sqrt{16-10\sqrt{2}}}{2}$$

Now solve the equation x=\frac{0±2i\sqrt{16-10\sqrt{2}}}{4} when ± is minus.

$$x=-\frac{i\sqrt{16-10\sqrt{2}}}{2}$$

The equation is now solved.

$$x=\frac{i\sqrt{16-10\sqrt{2}}}{2} x=-\frac{i\sqrt{16-10\sqrt{2}}}{2}$$