Solve 53x^2+3102x+40>0

$$53 x ^ { 2 } + 3102 x + 40 > 0$$

Solve for x

$$x\in \left(-\infty,\frac{-\sqrt{2403481}-1551}{53}\right)\cup \left(\frac{\sqrt{2403481}-1551}{53},\infty\right)$$

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.

$$53x^{2}+3102x+40=0$$

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, 3102 for b, and 40 for c in the quadratic formula.

$$x=\frac{-3102±\sqrt{3102^{2}-4\times 53\times 40}}{2\times 53}$$

Do the calculations.

$$x=\frac{-3102±2\sqrt{2403481}}{106}$$

Solve the equation x=\frac{-3102±2\sqrt{2403481}}{106} when ± is plus and when ± is minus.

$$x=\frac{\sqrt{2403481}-1551}{53} x=\frac{-\sqrt{2403481}-1551}{53}$$

Rewrite the inequality by using the obtained solutions.

$$53\left(x-\frac{\sqrt{2403481}-1551}{53}\right)\left(x-\frac{-\sqrt{2403481}-1551}{53}\right)>0$$

For the product to be positive, x-\frac{\sqrt{2403481}-1551}{53} and x-\frac{-\sqrt{2403481}-1551}{53} have to be both negative or both positive. Consider the case when x-\frac{\sqrt{2403481}-1551}{53} and x-\frac{-\sqrt{2403481}-1551}{53} are both negative.

$$x-\frac{\sqrt{2403481}-1551}{53}<0 x-\frac{-\sqrt{2403481}-1551}{53}<0$$

The solution satisfying both inequalities is x<\frac{-\sqrt{2403481}-1551}{53}.

$$x<\frac{-\sqrt{2403481}-1551}{53}$$

Consider the case when x-\frac{\sqrt{2403481}-1551}{53} and x-\frac{-\sqrt{2403481}-1551}{53} are both positive.

$$x-\frac{-\sqrt{2403481}-1551}{53}>0 x-\frac{\sqrt{2403481}-1551}{53}>0$$

The solution satisfying both inequalities is x>\frac{\sqrt{2403481}-1551}{53}.

$$x>\frac{\sqrt{2403481}-1551}{53}$$

The final solution is the union of the obtained solutions.

$$x<\frac{-\sqrt{2403481}-1551}{53}\text{; }x>\frac{\sqrt{2403481}-1551}{53}$$