the graph of y = ax^2 + bx + c is shown below. determine the solution set of 0 = ax^2 + bx + c.

Check the picture below.

something noteworthy to look at is that the graph doesn't cross the x-axis at -2, it simply comes down to it, touches it and it goes back up, it simply bounces off the x-axis, whenever that happens, that zero/solution/root has an even multiplicity.

when 0 = ax² + bx + c, we notice that y = 0, and for the graph that happens there, at x = -2, but that solution has an even multiplicity, and since the equation is a 2nd degree polynomial, thus x = -2 is there twice, namely

x = -2

x - 2 = 0

(x - 2)² = 0   < multiplicity of 2.

step-by-step explanation: