Solve (x^2-1)^2-18(x^2-1)+45=0

$$( x ^ { 2 } - 1 ) ^ { 2 } - 18 ( x ^ { 2 } - 1 ) + 45 = 0$$

Solve for x

$$x=-4$$

$$x=4$$

$$x=2$$

$$x=-2$$

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}.

$$\left(x^{2}\right)^{2}-2x^{2}+1-18\left(x^{2}-1\right)+45=0$$

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

$$x^{4}-2x^{2}+1-18\left(x^{2}-1\right)+45=0$$

Use the distributive property to multiply -18 by x^{2}-1.

$$x^{4}-2x^{2}+1-18x^{2}+18+45=0$$

Combine -2x^{2} and -18x^{2} to get -20x^{2}.

$$x^{4}-20x^{2}+1+18+45=0$$

Add 1 and 18 to get 19.

$$x^{4}-20x^{2}+19+45=0$$

Add 19 and 45 to get 64.

$$x^{4}-20x^{2}+64=0$$

Substitute t for x^{2}.

$$t^{2}-20t+64=0$$

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, -20 for b, and 64 for c in the quadratic formula.

$$t=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 1\times 64}}{2}$$

Do the calculations.

$$t=\frac{20±12}{2}$$

Solve the equation t=\frac{20±12}{2} when ± is plus and when ± is minus.

$$t=16 t=4$$

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

$$x=4 x=-4 x=2 x=-2$$