Answer
The value of x for the inequality $3{{x}^{2}}>2x+5$ is $\left( -\infty ,-1 \right)\cup \left( \frac{5}{3},\infty \right)$.
Work Step by Step
Solve:
$\begin{align}
& 3{{x}^{2}}>2x+5 \\
& 3{{x}^{2}}-2x-5>0 \\
& 3{{x}^{2}}+3x-5x-5>0 \\
& \left( x+1 \right)\left( 3x-5 \right)>0
\end{align}$
As the inequality is "greater than", either both of the factors should be positive or both should be negative.
Thus, apply the above condition to obtain:
$x\in \left( -\infty ,-1 \right)\cup \left( \frac{5}{3},\infty \right)$
The value of x for the inequality $3{{x}^{2}}>2x+5$ is $\left( -\infty ,-1 \right)\cup \left( \frac{5}{3},\infty \right)$.
