Solve 53x^2+5x-12<0

$$53 x ^ { 2 } + 5 x - 12 < 0$$

Solve for x

$$x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right)$$

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.

$$53x^{2}+5x-12=0$$

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

$$x=\frac{-5±\sqrt{5^{2}-4\times 53\left(-12\right)}}{2\times 53}$$

Do the calculations.

$$x=\frac{-5±\sqrt{2569}}{106}$$

Solve the equation x=\frac{-5±\sqrt{2569}}{106} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{2569}-5}{106} x=\frac{-\sqrt{2569}-5}{106}$$

Rewrite the inequality by using the obtained solutions.

$$53\left(x-\frac{\sqrt{2569}-5}{106}\right)\left(x-\frac{-\sqrt{2569}-5}{106}\right)<0$$

For the product to be negative, x-\frac{\sqrt{2569}-5}{106} and x-\frac{-\sqrt{2569}-5}{106} have to be of the opposite signs. Consider the case when x-\frac{\sqrt{2569}-5}{106} is positive and x-\frac{-\sqrt{2569}-5}{106} is negative.

$$x-\frac{\sqrt{2569}-5}{106}>0 x-\frac{-\sqrt{2569}-5}{106}<0$$

This is false for any x.

$$x\in \emptyset$$

Consider the case when x-\frac{-\sqrt{2569}-5}{106} is positive and x-\frac{\sqrt{2569}-5}{106} is negative.

$$x-\frac{-\sqrt{2569}-5}{106}>0 x-\frac{\sqrt{2569}-5}{106}<0$$

The solution satisfying both inequalities is x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right).

$$x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right)$$

The final solution is the union of the obtained solutions.

$$x\in \left(\frac{-\sqrt{2569}-5}{106},\frac{\sqrt{2569}-5}{106}\right)$$