Solve 62x^2+3x-1<0

$$62 x ^ { 2 } + 3 x - 1 < 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$

Steps Using the Quadratic Formula

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.

$$62x^{2}+3x-1=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 62\left(-1\right)}}{2\times 62}$$

Do the calculations.

$$x=\frac{-3±\sqrt{257}}{124}$$

Solve the equation x=\frac{-3±\sqrt{257}}{124} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{257}-3}{124} x=\frac{-\sqrt{257}-3}{124}$$

Rewrite the inequality by using the obtained solutions.

$$62\left(x-\frac{\sqrt{257}-3}{124}\right)\left(x-\frac{-\sqrt{257}-3}{124}\right)<0$$

For the product to be negative, x-\frac{\sqrt{257}-3}{124} and x-\frac{-\sqrt{257}-3}{124} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{257}-3}{124} is positive and x-\frac{-\sqrt{257}-3}{124} is negative.

$$x-\frac{\sqrt{257}-3}{124}>0 x-\frac{-\sqrt{257}-3}{124}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{257}-3}{124} is positive and x-\frac{\sqrt{257}-3}{124} is negative.

$$x-\frac{-\sqrt{257}-3}{124}>0 x-\frac{\sqrt{257}-3}{124}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right).

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{257}-3}{124},\frac{\sqrt{257}-3}{124}\right)$$