$3 \exponential{x}{2} - 10 x + 3 > 0 $
Steps Using the Quadratic Formula
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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(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\times 3}}{2\times 3}
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 3 for a, -10 for b, and 3 for c in the quadratic formula.
Solve the equation x=\frac{10±8}{6} when ± is plus and when ± is minus.
3\left(x-3\right)\left(x-\frac{1}{3}\right)>0
Rewrite the inequality by using the obtained solutions.
For the product to be positive, x-3 and x-\frac{1}{3} have to be both negative or both positive. Consider the case when x-3 and x-\frac{1}{3} are both negative.
The solution satisfying both inequalities is x<\frac{1}{3}.
Consider the case when x-3 and x-\frac{1}{3} are both positive.
The solution satisfying both inequalities is x>3.
x<\frac{1}{3}\text{; }x>3
The final solution is the union of the obtained solutions.