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

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

Solve for x

$$x\in \left(-\infty,-\frac{1}{3}\right)\cup \left(1,\infty\right)$$

Steps Using the Quadratic Formula

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

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

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.

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

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, -2 for b, and -1 for c in the quadratic formula.

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

Do the calculations.

$$x=\frac{2±4}{6}$$

Solve the equation x=\frac{2±4}{6} when ± is plus and when ± is minus.

$$x=1 x=-\frac{1}{3}$$

Rewrite the inequality by using the obtained solutions.

$$3\left(x-1\right)\left(x+\frac{1}{3}\right)>0$$

For the product to be positive, x-1 and x+\frac{1}{3} have to be both negative or both positive. Consider the case when x-1 and x+\frac{1}{3} are both negative.

$$x-1<0 x+\frac{1}{3}<0$$

The solution satisfying both inequalities is x<-\frac{1}{3}.

$$x<-\frac{1}{3}$$

Consider the case when x-1 and x+\frac{1}{3} are both positive.

$$x+\frac{1}{3}>0 x-1>0$$

The solution satisfying both inequalities is x>1.

$$x>1$$

The final solution is the union of the obtained solutions.

$$x<-\frac{1}{3}\text{; }x>1$$