Solve (x^2-8)^2+4(x^2-8)-5=0

$$( x ^ { 2 } - 8 ) ^ { 2 } + 4 ( x ^ { 2 } - 8 ) - 5 = 0$$

Solve for x

$$x=-3$$

$$x=3$$

$$x=\sqrt{3}\approx 1.732050808$$

$$x=-\sqrt{3}\approx -1.732050808$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x^{2}-8\right)^{2}.

$$\left(x^{2}\right)^{2}-16x^{2}+64+4\left(x^{2}-8\right)-5=0$$

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

$$x^{4}-16x^{2}+64+4\left(x^{2}-8\right)-5=0$$

Use the distributive property to multiply 4 by x^{2}-8.

$$x^{4}-16x^{2}+64+4x^{2}-32-5=0$$

Combine -16x^{2} and 4x^{2} to get -12x^{2}.

$$x^{4}-12x^{2}+64-32-5=0$$

Subtract 32 from 64 to get 32.

$$x^{4}-12x^{2}+32-5=0$$

Subtract 5 from 32 to get 27.

$$x^{4}-12x^{2}+27=0$$

Substitute t for x^{2}.

$$t^{2}-12t+27=0$$

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, -12 for b, and 27 for c in the quadratic formula.

$$t=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 1\times 27}}{2}$$

Do the calculations.

$$t=\frac{12±6}{2}$$

Solve the equation t=\frac{12±6}{2} when ± is plus and when ± is minus.

$$t=9 t=3$$

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

$$x=3 x=-3 x=\sqrt{3} x=-\sqrt{3}$$