Solve 5.3x^2+5x-12<0

$$5.3 x ^ { 2 } + 5 x - 12 < 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{6985}-25}{53},\frac{\sqrt{6985}-25}{53}\right)$$

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.

$$5.3x^{2}+5x-12=0$$

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 5.3 for a, 5 for b, and -12 for c in the quadratic formula.

$$x=\frac{-5±\sqrt{5^{2}-4\times 5.3\left(-12\right)}}{2\times 5.3}$$

Do the calculations.

$$x=\frac{-5±\frac{1}{5}\sqrt{6985}}{10.6}$$

Solve the equation x=\frac{-5±\frac{1}{5}\sqrt{6985}}{10.6} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{6985}-25}{53} x=\frac{-\sqrt{6985}-25}{53}$$

Rewrite the inequality by using the obtained solutions.

$$5.3\left(x-\frac{\sqrt{6985}-25}{53}\right)\left(x-\frac{-\sqrt{6985}-25}{53}\right)<0$$

For the product to be negative, x-\frac{\sqrt{6985}-25}{53} and x-\frac{-\sqrt{6985}-25}{53} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{6985}-25}{53} is positive and x-\frac{-\sqrt{6985}-25}{53} is negative.

$$x-\frac{\sqrt{6985}-25}{53}>0 x-\frac{-\sqrt{6985}-25}{53}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{6985}-25}{53} is positive and x-\frac{\sqrt{6985}-25}{53} is negative.

$$x-\frac{-\sqrt{6985}-25}{53}>0 x-\frac{\sqrt{6985}-25}{53}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{6985}-25}{53},\frac{\sqrt{6985}-25}{53}\right).

$$x\in \left(\frac{-\sqrt{6985}-25}{53},\frac{\sqrt{6985}-25}{53}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{6985}-25}{53},\frac{\sqrt{6985}-25}{53}\right)$$