Part 1

Imagine LHS is one graph, RHS is one graph.

So you always want to rewrite the equation such that LHS is the graph you have.

2x^2 + 3x - 8 = 0

So Y = 2x^2 + 3x - 8 is the LHS.
The drawn graph is Y = -2x^2 - 3x + 8, which is the negative version of the equation they are asking for, i.e. y = - (2×^2 + 3x - 8)
So you can imagine that u multiply both sides by -1.

Y = 0 is the RHS.
Multiplied by -1 is still 0.
Y= 0 drawn as a line is exactly the x-axis.

So we are looking for when the graph intersects the x-axis. That is the answer. (X = -2.9 or 1.4)

Part 2

Same thing, manipulate this equation so that left hand side = the graph we have.

-2x^2 - 3x + 5 = 0
Add 3 to both sides.
- 2x^2 - 3x + 8 = 3
So we need to draw a line that is y = 3.

And the answer is where this line intersects our graph. (X = -2.5 or 1)