## Precalculus (10th Edition)

$x$ is in the inverval: $(-0.5,3)$.
$2x^2<5x+3$ I subtract $(5x+3)$ from both sides: $2x^2-5x-3<0.$ The coefficient of $x^2$ is $2$, hence the coefficient of the xs in the multipliers should be $2$ and $1$. The coefficient of $x^0$ is -3, hence the coefficient of the $x^0$s in the multipliers should be $-3$ and $1$ or $3$ and $-1$. Hence the following combination is good: $(2x+1)(x-3)<0$. Therefore $x$ is in the inverval: $(-0.5,3)$ according to the graph.