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

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

Solve for a

$$a\in \left(-\infty,\frac{3-\sqrt{17}}{4}\right)\cup \left(\frac{\sqrt{17}+3}{4},\infty\right)$$

Steps Using the Quadratic Formula

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

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

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.

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

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

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

Do the calculations.

$$a=\frac{3±\sqrt{17}}{4}$$

Solve the equation a=\frac{3±\sqrt{17}}{4} when ± is plus and when ± is minus.

$$a=\frac{\sqrt{17}+3}{4} a=\frac{3-\sqrt{17}}{4}$$

Rewrite the inequality by using the obtained solutions.

$$2\left(a-\frac{\sqrt{17}+3}{4}\right)\left(a-\frac{3-\sqrt{17}}{4}\right)>0$$

For the product to be positive, a-\frac{\sqrt{17}+3}{4} and a-\frac{3-\sqrt{17}}{4} have to be both negative or both positive. Consider the case when a-\frac{\sqrt{17}+3}{4} and a-\frac{3-\sqrt{17}}{4} are both negative.

$$a-\frac{\sqrt{17}+3}{4}<0 a-\frac{3-\sqrt{17}}{4}<0$$

The solution satisfying both inequalities is a<\frac{3-\sqrt{17}}{4}.

$$a<\frac{3-\sqrt{17}}{4}$$

Consider the case when a-\frac{\sqrt{17}+3}{4} and a-\frac{3-\sqrt{17}}{4} are both positive.

$$a-\frac{3-\sqrt{17}}{4}>0 a-\frac{\sqrt{17}+3}{4}>0$$

The solution satisfying both inequalities is a>\frac{\sqrt{17}+3}{4}.

$$a>\frac{\sqrt{17}+3}{4}$$

The final solution is the union of the obtained solutions.

$$a<\frac{3-\sqrt{17}}{4}\text{; }a>\frac{\sqrt{17}+3}{4}$$