Solve (2x^2+2)^2-2(2x^2+2)-8=0

$$( 2 x ^ { 2 } + 2 ) ^ { 2 } - 2 ( 2 x ^ { 2 } + 2 ) - 8 = 0$$

Solve for x (complex solution)

$$x=1$$

$$x=-1$$

$$x=-\sqrt{2}i\approx -0-1.414213562i$$

$$x=\sqrt{2}i\approx 1.414213562i$$

Steps Using the Quadratic Formula

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

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

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

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

Subtract 4 from 4 to get 0.

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

Substitute t for x^{2}.

$$4t^{2}+4t-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, 4 for b, and -8 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-4±12}{8}$$

Solve the equation t=\frac{-4±12}{8} when ± is plus and when ± is minus.

$$t=1 t=-2$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for each t.

$$x=-1 x=1 x=-\sqrt{2}i x=\sqrt{2}i$$

Solve for x

$$x=-1$$

$$x=1$$

Steps Using the Quadratic Formula

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

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

To raise a power to another power, multiply the exponents. Multiply 2 and 2 to get 4.

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

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

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

Combine 8x^{2} and -4x^{2} to get 4x^{2}.

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

Subtract 4 from 4 to get 0.

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

Substitute t for x^{2}.

$$4t^{2}+4t-8=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. Substitute 4 for a, 4 for b, and -8 for c in the quadratic formula.

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

Do the calculations.

$$t=\frac{-4±12}{8}$$

Solve the equation t=\frac{-4±12}{8} when ± is plus and when ± is minus.

$$t=1 t=-2$$

Since x=t^{2}, the solutions are obtained by evaluating x=±\sqrt{t} for positive t.

$$x=1 x=-1$$