Solve 6.2x^2+3x-1<0

$$6.2 x ^ { 2 } + 3 x - 1 < 0$$

Solve for x

$$x\in \left(\frac{-13\sqrt{5}-15}{62},\frac{13\sqrt{5}-15}{62}\right)$$

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.

$$6.2x^{2}+3x-1=0$$

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

$$x=\frac{-3±\sqrt{3^{2}-4\times 6.2\left(-1\right)}}{2\times 6.2}$$

Do the calculations.

$$x=\frac{-3±\frac{13}{5}\sqrt{5}}{12.4}$$

Solve the equation x=\frac{-3±\frac{13}{5}\sqrt{5}}{12.4} when ± is plus and when ± is minus.

$$x=\frac{13\sqrt{5}-15}{62} x=\frac{-13\sqrt{5}-15}{62}$$

Rewrite the inequality by using the obtained solutions.

$$6.2\left(x-\frac{13\sqrt{5}-15}{62}\right)\left(x-\frac{-13\sqrt{5}-15}{62}\right)<0$$

For the product to be negative, x-\frac{13\sqrt{5}-15}{62} and x-\frac{-13\sqrt{5}-15}{62} have to be of the opposite signs. Consider the case when x-\frac{13\sqrt{5}-15}{62} is positive and x-\frac{-13\sqrt{5}-15}{62} is negative.

$$x-\frac{13\sqrt{5}-15}{62}>0 x-\frac{-13\sqrt{5}-15}{62}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-13\sqrt{5}-15}{62} is positive and x-\frac{13\sqrt{5}-15}{62} is negative.

$$x-\frac{-13\sqrt{5}-15}{62}>0 x-\frac{13\sqrt{5}-15}{62}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-13\sqrt{5}-15}{62},\frac{13\sqrt{5}-15}{62}\right).

$$x\in \left(\frac{-13\sqrt{5}-15}{62},\frac{13\sqrt{5}-15}{62}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-13\sqrt{5}-15}{62},\frac{13\sqrt{5}-15}{62}\right)$$