Solve 5.2x^2-5x-3<0Hxrightarrow

$$5.2 x ^ { 2 } - 5 x - 3 < 0 H x \rightarrow$$

Solve for x

$$x\in \left(\frac{25-\sqrt{2185}}{52},\frac{\sqrt{2185}+25}{52}\right)$$

Steps Using the Quadratic Formula

Anything times zero gives zero.

$$5.2x^{2}-5x-3<0$$

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.

$$5.2x^{2}-5x-3=0$$

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 5.2 for a, -5 for b, and -3 for c in the quadratic formula.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 5.2\left(-3\right)}}{2\times 5.2}$$

Do the calculations.

$$x=\frac{5±\frac{1}{5}\sqrt{2185}}{10.4}$$

Solve the equation x=\frac{5±\frac{1}{5}\sqrt{2185}}{10.4} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{2185}+25}{52} x=\frac{25-\sqrt{2185}}{52}$$

Rewrite the inequality by using the obtained solutions.

$$5.2\left(x-\frac{\sqrt{2185}+25}{52}\right)\left(x-\frac{25-\sqrt{2185}}{52}\right)<0$$

For the product to be negative, x-\frac{\sqrt{2185}+25}{52} and x-\frac{25-\sqrt{2185}}{52} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{2185}+25}{52} is positive and x-\frac{25-\sqrt{2185}}{52} is negative.

$$x-\frac{\sqrt{2185}+25}{52}>0 x-\frac{25-\sqrt{2185}}{52}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{25-\sqrt{2185}}{52} is positive and x-\frac{\sqrt{2185}+25}{52} is negative.

$$x-\frac{25-\sqrt{2185}}{52}>0 x-\frac{\sqrt{2185}+25}{52}<0$$

The solution satisfying both inequalities is x\in \left(\frac{25-\sqrt{2185}}{52},\frac{\sqrt{2185}+25}{52}\right).

$$x\in \left(\frac{25-\sqrt{2185}}{52},\frac{\sqrt{2185}+25}{52}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{25-\sqrt{2185}}{52},\frac{\sqrt{2185}+25}{52}\right)$$