Use the completing the square method. -4s^2 + 2s + 6 = 0

To use the completing the square method, we first need to isolate the squared term and the linear term on one side of the equation:

-4s^2 + 2s + 6 = 0
-4s^2 + 2s = -6

Next, we divide the coefficient of the linear term by 2 and square the result:

b = 2
b/2 = 2/2 = 1
(1)^2 = 1

Now, we add and subtract this squared value inside the parentheses:

-4s^2 + 2s + 1 - 1 = -6
-4s^2 + 2s + 1 = -5

Now, we rewrite the equation as a perfect square trinomial:

(-2s + 1)^2 = -5

To solve for s, we take the square root of both sides:

-2s + 1 = ±√(-5)
-2s + 1 = ±i√5

Now, we isolate s:

-2s = -1 ± i√5
s = (1 ± i√5) / 2

Therefore, the solutions to the equation -4s^2 + 2s + 6 = 0 are s = (1 + i√5) / 2 and s = (1 - i√5) / 2.

Step 1: Move the constant term (6) to the right side of the equation.

-4s^2 + 2s = -6

Step 2: Divide both sides of the equation by the coefficient of the squared term (-4).

s^2 - (1/2)s = 3/2

Step 3: Add the square of half the coefficient of the linear term ((1/4)^2 = 1/16) to both sides of the equation.

s^2 - (1/2)s + 1/16 = 3/2 + 1/16

Step 4: Factor the left side of the equation as a perfect square trinomial.

(s - 1/4)^2 = 25/16

Step 5: Take the square root of both sides of the equation.

s - 1/4 = ±5/4

Step 6: Solve for s.

s = 1/4 ± 5/4
s = 3/2 or s = -1/2