For the quadratic function y = ax 2+bx+c with two intersections with the x axis, the distance between these two intersections is d = | δ 0.5/a |, and the function in this question must have two intersections with the x axis. So if you put it in the formula, you will get

d=|(a^2-4(a-3))^0.5|

=(a^2-4a+ 12)^0.5

The problem is transformed into finding the minimum value of the function y = a 2-4a+ 12.

The minimum value of this function is: y (min) = (a-2) 2+8.

=8

So the minimum required is 8 0.5 = 2 times the root number 2.

I made a mistake myself, sorry! !

In Word, you can use "Insert" and then select "Object". There is a formula editor in which you can add mathematical formulas.

Ok, 1 1

Note that d= (discriminant) 0.5, so open the root sign.