Solve -2(x^2-1)-(2x-1)<0

$$- 2 ( x ^ { 2 } - 1 ) - ( 2 x - 1 ) < 0$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{7}-1}{2}\right)\cup \left(\frac{\sqrt{7}-1}{2},\infty\right)$$

Steps Using the Quadratic Formula

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

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

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

$$-2x^{2}+2-2x+1<0$$

Add 2 and 1 to get 3.

$$-2x^{2}+3-2x<0$$

Multiply the inequality by -1 to make the coefficient of the highest power in -2x^{2}+3-2x positive. Since -1 is negative, the inequality direction is changed.

$$2x^{2}-3+2x>0$$

To solve the inequality, factor the left hand side. Quadratic polynomial can be factored using the transformation ax^{2}+bx+c=a\left(x-x_{1}\right)\left(x-x_{2}\right), where x_{1} and x_{2} are the solutions of the quadratic equation ax^{2}+bx+c=0.

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

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, 2 for b, and -3 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{-2±2\sqrt{7}}{4}$$

Solve the equation x=\frac{-2±2\sqrt{7}}{4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{7}-1}{2} x=\frac{-\sqrt{7}-1}{2}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(x-\frac{\sqrt{7}-1}{2}\right)\left(x-\frac{-\sqrt{7}-1}{2}\right)>0$$

For the product to be positive, x-\frac{\sqrt{7}-1}{2} and x-\frac{-\sqrt{7}-1}{2} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{7}-1}{2} and x-\frac{-\sqrt{7}-1}{2} are both negative.

$$x-\frac{\sqrt{7}-1}{2}<0 x-\frac{-\sqrt{7}-1}{2}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{7}-1}{2}.

$$x<\frac{-\sqrt{7}-1}{2}$$

Consider the case when x-\frac{\sqrt{7}-1}{2} and x-\frac{-\sqrt{7}-1}{2} are both positive.

$$x-\frac{-\sqrt{7}-1}{2}>0 x-\frac{\sqrt{7}-1}{2}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{7}-1}{2}.

$$x>\frac{\sqrt{7}-1}{2}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{7}-1}{2}\text{; }x>\frac{\sqrt{7}-1}{2}$$