# How do you solve 2p^2 - 7p + 3 > 0?

Jun 6, 2015

$f \left(p\right) = 2 {p}^{2} - 7 p + 3 = \left(2 p - 1\right) \left(p - 3\right)$

$f \left(p\right) = 0$ when $p = \frac{1}{2}$ or $p = 3$
So the graph of $f \left(p\right)$ crosses the horizontal axis at $\left(\frac{1}{2} , 0\right)$ and $\left(3 , 0\right)$.

For large positive or negative values of $p$, the $2 {p}^{2}$ term will dominate, resulting in $f \left(p\right) > 0$

Hence $f \left(p\right) > 0$ when $p < \frac{1}{2}$ or $p > 3$

graph{2x^2-7x+3 [-10, 10, -5, 5]}