The parabola has equation

#y=ax^2+bx+c#

and we have to find three parameters to determine it: #a, b, c#.

To find them we have to use the three given points that are

#(-6, 0), (5,0), (0, 3)#. The zeros are because the points are intercepts, it means that in those points they cross or the #y# axes (for the first two) or the #x# axes (for the last one).

We can substitute the values of the points in the equation

#0=a*(-6)^2+b*(-6)+c#

#0=a*5^2+b*5+c#

#3=a*0^2+b*0+c#

I do the calculations and have

#0=36a-6b+c#

#0=25a+5b+c#

#3=c#

We are lucky! From the third equation we have the value of #c# that we can use in the first two, so we have

#0=36a-6b+3#

#0=25a+5b+3#

#3=c#

We find #a# from the first equation

#0=36a-6b+3#

#36a=6b-3#

#a=(6b-3)/36=b/6-1/12#

and we substitute this value in the second equation

#0=25a+5b+3#

#0=25(b/6-1/12)+5b+3#

#0=25/6b+5b+3-25/12#

#0=(25+30)/6b+(36-25)/12#

#0=55/6b+11/12#

#55/6b=-11/12#

#b=-1/10#.

And finally I use this value of #b# in the previous equation

#a=b/6-1/12#

#a=-1/10*1/6-1/12=-1/60-1/12=-1/60-5/60=-6/60=-1/10#

Our three numbers are #a=-1/10, b=-1/10, c=3# and the parabola is

#y=-1/10x^2-1/10x+3#. We can verify looking if the plot pass for the three points #(-6, 0), (5,0), (0, 3)#.

graph{y=-1/10x^2-1/10x+3 [-10, 10, -5, 5]}