Solve x^2=2+(17-2x^2-2*4sqrt{2}(17-2x)*frac{sqrt{2}}{2})

$$x ^ { 2 } = 2 + ( 17 - 2 x ^ { 2 } - 2 \cdot 4 \sqrt { 2 } ( 17 - 2 x ) \cdot \frac { \sqrt { 2 } } { 2 } )$$

Solve for x (complex solution)

$$x=\frac{8+\sqrt{287}i}{3}\approx 2.666666667+5.647024782i$$

$$x=\frac{-\sqrt{287}i+8}{3}\approx 2.666666667-5.647024782i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

$$2x^{2}=4+2\left(17-2x^{2}-2\times 4\sqrt{2}\left(17-2x\right)\times \frac{\sqrt{2}}{2}\right)$$

Multiply 2 and 4 to get 8.

$$2x^{2}=4+2\left(17-2x^{2}-8\sqrt{2}\left(17-2x\right)\times \frac{\sqrt{2}}{2}\right)$$

Cancel out 2, the greatest common factor in 8 and 2.

$$2x^{2}=4+2\left(17-2x^{2}-4\sqrt{2}\sqrt{2}\left(17-2x\right)\right)$$

Use the distributive property to multiply 4\sqrt{2}\sqrt{2} by 17-2x.

$$2x^{2}=4+2\left(17-2x^{2}-\left(17\times 4\sqrt{2}\sqrt{2}-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2x^{2}=4+2\left(17-2x^{2}-\left(17\times 4\times 2-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

Multiply 17 and 4 to get 68.

$$2x^{2}=4+2\left(17-2x^{2}-\left(68\times 2-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

Multiply 68 and 2 to get 136.

$$2x^{2}=4+2\left(17-2x^{2}-\left(136-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

The square of \sqrt{2} is 2.

$$2x^{2}=4+2\left(17-2x^{2}-\left(136-8x\times 2\right)\right)$$

Multiply -8 and 2 to get -16.

$$2x^{2}=4+2\left(17-2x^{2}-\left(136-16x\right)\right)$$

To find the opposite of 136-16x, find the opposite of each term.

$$2x^{2}=4+2\left(17-2x^{2}-136+16x\right)$$

Subtract 136 from 17 to get -119.

$$2x^{2}=4+2\left(-119-2x^{2}+16x\right)$$

Use the distributive property to multiply 2 by -119-2x^{2}+16x.

$$2x^{2}=4-238-4x^{2}+32x$$

Subtract 238 from 4 to get -234.

$$2x^{2}=-234-4x^{2}+32x$$

Subtract -234 from both sides.

$$2x^{2}-\left(-234\right)=-4x^{2}+32x$$

The opposite of -234 is 234.

$$2x^{2}+234=-4x^{2}+32x$$

Add 4x^{2} to both sides.

$$2x^{2}+234+4x^{2}=32x$$

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}+234=32x$$

Subtract 32x from both sides.

$$6x^{2}+234-32x=0$$

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.

$$6x^{2}-32x+234=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -32 for b, and 234 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-32\right)±\sqrt{\left(-32\right)^{2}-4\times 6\times 234}}{2\times 6}$$

Square -32.

$$x=\frac{-\left(-32\right)±\sqrt{1024-4\times 6\times 234}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-32\right)±\sqrt{1024-24\times 234}}{2\times 6}$$

Multiply -24 times 234.

$$x=\frac{-\left(-32\right)±\sqrt{1024-5616}}{2\times 6}$$

Add 1024 to -5616.

$$x=\frac{-\left(-32\right)±\sqrt{-4592}}{2\times 6}$$

Take the square root of -4592.

$$x=\frac{-\left(-32\right)±4\sqrt{287}i}{2\times 6}$$

The opposite of -32 is 32.

$$x=\frac{32±4\sqrt{287}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{32±4\sqrt{287}i}{12}$$

Now solve the equation x=\frac{32±4\sqrt{287}i}{12} when ± is plus. Add 32 to 4i\sqrt{287}.

$$x=\frac{32+4\sqrt{287}i}{12}$$

Divide 32+4i\sqrt{287} by 12.

$$x=\frac{8+\sqrt{287}i}{3}$$

Now solve the equation x=\frac{32±4\sqrt{287}i}{12} when ± is minus. Subtract 4i\sqrt{287} from 32.

$$x=\frac{-4\sqrt{287}i+32}{12}$$

Divide 32-4i\sqrt{287} by 12.

$$x=\frac{-\sqrt{287}i+8}{3}$$

The equation is now solved.

$$x=\frac{8+\sqrt{287}i}{3} x=\frac{-\sqrt{287}i+8}{3}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

$$2x^{2}=4+2\left(17-2x^{2}-2\times 4\sqrt{2}\left(17-2x\right)\times \frac{\sqrt{2}}{2}\right)$$

Multiply 2 and 4 to get 8.

$$2x^{2}=4+2\left(17-2x^{2}-8\sqrt{2}\left(17-2x\right)\times \frac{\sqrt{2}}{2}\right)$$

Cancel out 2, the greatest common factor in 8 and 2.

$$2x^{2}=4+2\left(17-2x^{2}-4\sqrt{2}\sqrt{2}\left(17-2x\right)\right)$$

Use the distributive property to multiply 4\sqrt{2}\sqrt{2} by 17-2x.

$$2x^{2}=4+2\left(17-2x^{2}-\left(17\times 4\sqrt{2}\sqrt{2}-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$2x^{2}=4+2\left(17-2x^{2}-\left(17\times 4\times 2-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

Multiply 17 and 4 to get 68.

$$2x^{2}=4+2\left(17-2x^{2}-\left(68\times 2-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

Multiply 68 and 2 to get 136.

$$2x^{2}=4+2\left(17-2x^{2}-\left(136-8x\left(\sqrt{2}\right)^{2}\right)\right)$$

The square of \sqrt{2} is 2.

$$2x^{2}=4+2\left(17-2x^{2}-\left(136-8x\times 2\right)\right)$$

Multiply -8 and 2 to get -16.

$$2x^{2}=4+2\left(17-2x^{2}-\left(136-16x\right)\right)$$

To find the opposite of 136-16x, find the opposite of each term.

$$2x^{2}=4+2\left(17-2x^{2}-136+16x\right)$$

Subtract 136 from 17 to get -119.

$$2x^{2}=4+2\left(-119-2x^{2}+16x\right)$$

Use the distributive property to multiply 2 by -119-2x^{2}+16x.

$$2x^{2}=4-238-4x^{2}+32x$$

Subtract 238 from 4 to get -234.

$$2x^{2}=-234-4x^{2}+32x$$

Add 4x^{2} to both sides.

$$2x^{2}+4x^{2}=-234+32x$$

Combine 2x^{2} and 4x^{2} to get 6x^{2}.

$$6x^{2}=-234+32x$$

Subtract 32x from both sides.

$$6x^{2}-32x=-234$$

Divide both sides by 6.

$$\frac{6x^{2}-32x}{6}=-\frac{234}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\left(-\frac{32}{6}\right)x=-\frac{234}{6}$$

Reduce the fraction \frac{-32}{6} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{16}{3}x=-\frac{234}{6}$$

Divide -234 by 6.

$$x^{2}-\frac{16}{3}x=-39$$

Divide -\frac{16}{3}, the coefficient of the x term, by 2 to get -\frac{8}{3}. Then add the square of -\frac{8}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{16}{3}x+\left(-\frac{8}{3}\right)^{2}=-39+\left(-\frac{8}{3}\right)^{2}$$

Square -\frac{8}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=-39+\frac{64}{9}$$

Add -39 to \frac{64}{9}.

$$x^{2}-\frac{16}{3}x+\frac{64}{9}=-\frac{287}{9}$$

Factor x^{2}-\frac{16}{3}x+\frac{64}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{8}{3}\right)^{2}=-\frac{287}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{8}{3}\right)^{2}}=\sqrt{-\frac{287}{9}}$$

Simplify.

$$x-\frac{8}{3}=\frac{\sqrt{287}i}{3} x-\frac{8}{3}=-\frac{\sqrt{287}i}{3}$$

Add \frac{8}{3} to both sides of the equation.

$$x=\frac{8+\sqrt{287}i}{3} x=\frac{-\sqrt{287}i+8}{3}$$