Solve left(2x^2+2xright)^2-8x(x^2+1)=0

$${ \left(2 { x }^{ 2 } +2x \right) }^{ 2 } -8x( { x }^{ 2 } +1)=0$$

Solve for x (complex solution)

$$x=1$$

$$x=0$$

$$x=\frac{-\sqrt{7}i-1}{2}\approx -0.5-1.322875656i$$

$$x=\frac{-1+\sqrt{7}i}{2}\approx -0.5+1.322875656i$$

Steps Using the Quadratic Formula

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

$$4\left(x^{2}\right)^{2}+8x^{2}x+4x^{2}-8x\left(x^{2}+1\right)=0$$

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

$$4x^{4}+8x^{2}x+4x^{2}-8x\left(x^{2}+1\right)=0$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4x^{4}+8x^{3}+4x^{2}-8x\left(x^{2}+1\right)=0$$

Use the distributive property to multiply -8x by x^{2}+1.

$$4x^{4}+8x^{3}+4x^{2}-8x^{3}-8x=0$$

Combine 8x^{3} and -8x^{3} to get 0.

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

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

$$x=1$$

$$x=0$$

Steps Using the Quadratic Formula

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

$$4\left(x^{2}\right)^{2}+8x^{2}x+4x^{2}-8x\left(x^{2}+1\right)=0$$

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

$$4x^{4}+8x^{2}x+4x^{2}-8x\left(x^{2}+1\right)=0$$

To multiply powers of the same base, add their exponents. Add 2 and 1 to get 3.

$$4x^{4}+8x^{3}+4x^{2}-8x\left(x^{2}+1\right)=0$$

Use the distributive property to multiply -8x by x^{2}+1.

$$4x^{4}+8x^{3}+4x^{2}-8x^{3}-8x=0$$

Combine 8x^{3} and -8x^{3} to get 0.

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

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.

$$x=1 x=-1$$