I need help solving this Trinomial?

-6x^2 + 23x + 4
=-(6x^2-23x-4)
=-(6x+1)(x-4) answer//

-6x^2 + 23x + 4
= -6x^2 + 24x - x + 4
= (-6x^2 + 24x) - (x - 4)
= -6x(x - 4) - 1(x - 4)
= (x - 4)(-6x - 1)
= -(x - 4)(6x + 1)

to make it easier or maybe its just me.. i make 6x^2 positive so multiply the whole thing with a negative sign
6x^2 -23x -4=0
(6x + 1)(x - 4)=0
6x+1 = 0
6x = -1
x= -1/6
and
x = 4

x=[-23+&-(529+96)^1/2]/-12=-1/6 & 4
-6(x+1/6)(x-4)=(4-x)(6x+1)

Easy! -6x²+23x+4 is a trinomial with factors a, b, and c:
a=-6
b=23
c=4

Calculate ∆=b²-4ac=23²-4*(-6)*4=625
√∆=√625=25

The two zeros are x=(-b-√∆)/(2a) and x=(-b+√∆)/(2a)
ie x=(-23-25)/(-12) and x=(-23-25)/(-12)
ie x=4 and x=-1/6

Now, when you know the zeros in a trinomial (let's call them d an e), you can write the trinomial as a*(x-d)*(x-e), with so here it would be:

-6x²+23x+4=-6(x-4)(x+1/6)

- 6x² + 23x + 4 = 0
x² - 23/12x = 2/3 + (- 23/12)²
x² - 23/12x = 96/144 + 529/144
(x - 23/12)² = 625/144
x - 23/12 = +/- 25/12

= x - 23/12 - 25/12, = x - 48/12, = x - 4
= x - 23/12 + 25/12, = x + 2/12, = x + 1/6, = 6x + 1

Answer: - (x - 4)(6x + 1) are the factors.

Proof:
= - ([x - 4][6x + 1])
= - ([x * 6x] + [x * 1] + [- 4 * 6x] + [- 4 * 1])
= - (6x² + x - 24x - 4)
= - (6x² - 23x - 4)
= - 6x² + 23x + 4