It is known that $x\in [0,10]$. What is the possibility that $x^{2}+b > ax$ is true, when $a = 15.3$ and $b = 58.5$.
Is it correct to calculate this like that: $x^{2} + b$ is more than $ax$ when $ x^{2} + 58.5$ is more than $15.3x$, then calculate the quadratic equation and get $x_{1} = 7.8$ and $x_{2} = 7.5$. $7.5\leq x\leq 7.8$ is when $x^{2}+b > ax$ is not correct, therefore, the answer to this question is $$\frac{10-(7.8-7.5))}{10} = \frac{10-0.3}{10} = \frac{9.7}{10} = 0.97$$
$x^2+ 58.5> 15.3x$ is equivalent to $x^2- 15.3x+ 58.5> 0$. Completing the square, $x^2- 15.3x+ 58.5225- 58.5225+ 58.5= (x- 7.65)^2- 0.0225> 0$. That is the same as $(x- 7.65)^2> 0.0225$ which implies that either $x- 7.65> 0.15$, so that $x> 7.80$, or $x- 7.65< -0.15$, so that x< 7.5. You say that "it is known that $x\in [0, 10]$" but you don't say what the probability distribution is. I am going to assume a uniform distribution. The probability that x> 7.8 is $\frac{10- 7.8}{10- 0}= 0.22$. The probability that x< 7.5 is $\frac{7.5}{10}= 0.75$. The desired probability is 0.22+ 0.75= 0.95.