Solve -3x^2-x+1>0

$$- 3 x ^ { 2 } - x + 1 > 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{13}-1}{6},\frac{\sqrt{13}-1}{6}\right)$$

Steps Using the Quadratic Formula

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

$$3x^{2}+x-1<0$$

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.

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

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

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

Do the calculations.

$$x=\frac{-1±\sqrt{13}}{6}$$

Solve the equation x=\frac{-1±\sqrt{13}}{6} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{13}-1}{6} x=\frac{-\sqrt{13}-1}{6}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-\frac{\sqrt{13}-1}{6}\right)\left(x-\frac{-\sqrt{13}-1}{6}\right)<0$$

For the product to be negative, x-\frac{\sqrt{13}-1}{6} and x-\frac{-\sqrt{13}-1}{6} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{13}-1}{6} is positive and x-\frac{-\sqrt{13}-1}{6} is negative.

$$x-\frac{\sqrt{13}-1}{6}>0 x-\frac{-\sqrt{13}-1}{6}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{13}-1}{6} is positive and x-\frac{\sqrt{13}-1}{6} is negative.

$$x-\frac{-\sqrt{13}-1}{6}>0 x-\frac{\sqrt{13}-1}{6}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{13}-1}{6},\frac{\sqrt{13}-1}{6}\right).

$$x\in \left(\frac{-\sqrt{13}-1}{6},\frac{\sqrt{13}-1}{6}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{13}-1}{6},\frac{\sqrt{13}-1}{6}\right)$$