Solve (x^2-1)^2-(x^2-1)-6=0

$$( x ^ { 2 } - 1 ) ^ { 2 } - ( x ^ { 2 } - 1 ) - 6 = 0$$

Solve for x (complex solution)

$$x=2$$

$$x=-2$$

$$x=-i$$

$$x=i$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$\left(x^{2}\right)^{2}-2x^{2}+1-\left(x^{2}-1\right)-6=0$$

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

$$x^{4}-2x^{2}+1-\left(x^{2}-1\right)-6=0$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{4}-2x^{2}+1-x^{2}+1-6=0$$

Combine -2x^{2} and -x^{2} to get -3x^{2}.

$$x^{4}-3x^{2}+1+1-6=0$$

Add 1 and 1 to get 2.

$$x^{4}-3x^{2}+2-6=0$$

Subtract 6 from 2 to get -4.

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

Substitute t for x^{2}.

$$t^{2}-3t-4=0$$

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 1 for a, -3 for b, and -4 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$t=\frac{3±5}{2}$$

Solve the equation t=\frac{3±5}{2} when ± is plus and when ± is minus.

$$t=4 t=-1$$

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

$$x=-2 x=2 x=-i x=i$$

Solve for x

$$x=-2$$

$$x=2$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-1\right)^{2}.

$$\left(x^{2}\right)^{2}-2x^{2}+1-\left(x^{2}-1\right)-6=0$$

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

$$x^{4}-2x^{2}+1-\left(x^{2}-1\right)-6=0$$

To find the opposite of x^{2}-1, find the opposite of each term.

$$x^{4}-2x^{2}+1-x^{2}+1-6=0$$

Combine -2x^{2} and -x^{2} to get -3x^{2}.

$$x^{4}-3x^{2}+1+1-6=0$$

Add 1 and 1 to get 2.

$$x^{4}-3x^{2}+2-6=0$$

Subtract 6 from 2 to get -4.

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

Substitute t for x^{2}.

$$t^{2}-3t-4=0$$

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 1 for a, -3 for b, and -4 for c in the quadratic formula.

$$t=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\times 1\left(-4\right)}}{2}$$

Do the calculations.

$$t=\frac{3±5}{2}$$

Solve the equation t=\frac{3±5}{2} when ± is plus and when ± is minus.

$$t=4 t=-1$$

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

$$x=2 x=-2$$