Question 1060424
Easier to make the graph if you know the vertex and the axis intercepts.  Your parabola is concave down (see the coefficient on the leading term negative).


{{{y=-5(x^2-(2/5)x-3/5)}}}


{{{y=-5(x^2-(2/5)x+(2/20)^2-(2/20)^2-3/5)}}}------this step is Completing The Square.


{{{y=-5(x^2-(2/5)x+(1/10)^2-(1/10)^2-3/5)}}}


{{{y=-5((x-1/10)^2-1/100-60/100)}}}


{{{y=-5(x-1/10)^2+61/20}}}


The MAXIMUM value, vertex, is at  ( 1/10, 61/20 ), and you can solve the equation for the zeros of the graph.


{{{graph(300,300,-4,4,-4,4,-5x^2+2x+3)}}}