Solve -2x(x^3-3x-4)=0

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

Solve for x

$$x=\sqrt[3]{\sqrt{3}+2}+\sqrt[3]{2-\sqrt{3}}\approx 2.195823345$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply -2x by x^{3}-3x-4.

$$-2x^{4}+6x^{2}+8x=0$$

Substitute t for x^{2}.

$$-2t^{2}+6t+8=0$$

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 -2 for a, 6 for b, and 8 for c in the quadratic formula.

$$t=\frac{-6±\sqrt{6^{2}-4\left(-2\right)\times 8}}{-2\times 2}$$

Do the calculations.

$$t=\frac{-6±10}{-4}$$

Solve the equation t=\frac{-6±10}{-4} when ± is plus and when ± is minus.

$$t=-1 t=4$$

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

$$x=2 x=-2$$