Can You Show Me Step By Step How To Solve 2x^2-3x-2=0 In Quadratic Formula?

The quadratic formula gives the solution to
ax²+bx+c = 0

as
x = (-b ±√(b²-4ac))/(2a)    (this is "the quadratic formula")

You have
2x² -3x -2 = 0

So a=2, b=-3, c=-2.
We substitute these values into the quadratic formula.
x = (-b ±√(b²-4(2)c))/(2(2))    (we have put (2) where a was)
x = (-(-3) ±√((-3)²-4(2)c))/(2(2))    (we have put (-3) where b was)
x = (-(-3) ±√((-3)²-4(2)(-2)))/(2(2))    (we have put (-2) where c was)

Now, we evaluate according to the order of operations.
First, we evaluate parentheses. ((-3)²-4(2)(-2))
Within that, we evaluate exponents. (-3)² = 9, so the parentheses enclose (9 -4(2)(-2))
Then, we multiply (9 -8(-2)), (9 + 16)
Then, we add (25).

Now, we have
x = (-(-3) ±√(25))/(2(2))

Next, we do the root. √25 = 5
Now, we have

x = (-(-3) ±5)/(2(2))


Next, we do the multiplication inside each of the parentheses.
x = (3 ±5)/4

Now, parentheses again
x = 8/4    (when we use the + sign)
x = -2/4    (when we use the - sign)

Finally, we can perform the division
x = 2
x = -1/2

This is your solution.