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39x^{2}-156x+1=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.
x=\frac{-\left(-156\right)±\sqrt{\left(-156\right)^{2}-4\times 39\times 1}}{2\times 39}
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 39 for a, -156 for b, and 1 for c in the quadratic formula.
x=\frac{156±2\sqrt{6045}}{78}
Do the calculations.
x=\frac{\sqrt{6045}}{39}+2 x=-\frac{\sqrt{6045}}{39}+2
Solve the equation x=\frac{156±2\sqrt{6045}}{78} when ± is plus and when ± is minus.
39\left(x-\left(\frac{\sqrt{6045}}{39}+2\right)\right)\left(x-\left(-\frac{\sqrt{6045}}{39}+2\right)\right)>0
Rewrite the inequality by using the obtained solutions.
x-\left(\frac{\sqrt{6045}}{39}+2\right)<0 x-\left(-\frac{\sqrt{6045}}{39}+2\right)<0
For the product to be positive, x-\left(\frac{\sqrt{6045}}{39}+2\right) and x-\left(-\frac{\sqrt{6045}}{39}+2\right) have to be both negative or both positive. Consider the case when x-\left(\frac{\sqrt{6045}}{39}+2\right) and x-\left(-\frac{\sqrt{6045}}{39}+2\right) are both negative.
x<-\frac{\sqrt{6045}}{39}+2
The solution satisfying both inequalities is x<-\frac{\sqrt{6045}}{39}+2.
x-\left(-\frac{\sqrt{6045}}{39}+2\right)>0 x-\left(\frac{\sqrt{6045}}{39}+2\right)>0
Consider the case when x-\left(\frac{\sqrt{6045}}{39}+2\right) and x-\left(-\frac{\sqrt{6045}}{39}+2\right) are both positive.
x>\frac{\sqrt{6045}}{39}+2
The solution satisfying both inequalities is x>\frac{\sqrt{6045}}{39}+2.
x<-\frac{\sqrt{6045}}{39}+2\text{; }x>\frac{\sqrt{6045}}{39}+2
The final solution is the union of the obtained solutions.