Can you factor this equation

I'm trying to complete the square so I can graph this:

f(x) = -4x² - 6x + 2

I'm at:

f(x) = -4(x² + (3/2)x + (3/4)) + 5

So, now I'm trying to factor the equation in the parantheses and have no clue what it could be.
Did I make an error? Someone help!

...your work is incorrect !

f(x) = -4x² - 6x + 2

f(x) = - 4(x^2 + (3/2)x + ___) + 2

f(x) = - 4(x^2 + (3/2)x + (9/16)) + 2 + (9/4)

f(x) = - 4(x + (3/4))^2 + (17/4)

you can check by expanding and compare to original function.

qed

For any quadratic equation aX^2 +bX =C you can factor as follows

X = (-b + or - square root of [b^2 - 4ac])/2a
For -4x^2 - 6x +2
Take out common factor of -2 to get

-2( 2x^2 + 3x -1)

Now apply the formula to the expression in ( )

( -3 + or - square root( 9 - [4x2x(-1)})/2x2

(-3 + or - (square root 17)/4 = (-3 + or - [4.14])/4

the two values of x that satisfy the equation are -7.14/4 and 1.14/4

When you complete the square, factoring is given to you free, the second term in your factor is the square root of your third term...

x^2 + 6x + 9 for example, gives you (x + 3)(x + 3)

The issue with yours is that your complete square isn't right.

-4x² - 6x + 2
-4(x^2 + (3/2)x - 1/2)
-4(x^2 + (3/2)x) + 2 now get your 3rd polynomial term by halving 3/2 and squaring it
-4(x^2 + (3/2)x + 3/16) + 2 - 12/16
-4(x^2 + (3/2)x + 3/16) + 32/16 - 12/16
-4(x^2 + (3/2)x + 3/16) + 20/16
-4(x + 3/4)(x + 3/4) + 20/16

If you're attempting to complete the square, you made an error. The coefficient of the x term is 3/2, so the constant term must be half that, squared, which is (3/4)^2, or 9/16.

f(x) = -4(x^2 + (3/2)x + 9/16) + 9/4 + 2
f(x) = -4(x^2 + (3/2)x + 9/16) + 17/4
f(x) = -4(x + 3/4)^2 + 17/4