Answer
This statement is true.
Work Step by Step
The function $y=ax^2+bx+c$ is a quadratic polynomial function. The intercepts with the $x$ axis will occur when $y=0$. Putting $y=0$ we get
$$ax^2+bx+c=0$$ and the solutions for $x$ determine the intercepts with the $x$ axis.
Now this equation is a quadratic equation and its' solutions are
$$x_{1,2} = \frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Since $b^2-4ac\gt0$ its root will exist and it will be bigger that zero so there will be two different solutions of the equation:
$$x_1=\frac{-b+\sqrt{b^2-4ac}}{2a};\quad x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}$$
which means we will have two intercepts, one in $x_1$ and the other in $x_2$.
