I have to solve 4 - x - 3x^2, but isn't this problem backwards

Looks "backwards"

Turn it around.......Anytime the first term is negative, factor out a negative

4 - x - 3x^2

-3x^2 - x + 4

-1(3x^2 + x - 4).............Now factor........bring down the "-"

ANSWER............-(3x+4)(x-1)

yes, the problem is 'backwards' from what you are used to seeing
but when you consider the quadratic as
(a+b)^2

both ends are squares
a^2+2ab+b^2

so you solve it like you would if it was
4x^2-x-3

using the quadratic formula
the discriminant
(b^2-4ac)
1+48
49
+/-7

rest of q.f.
(-b+/-7)/2a

(1+/-7)/8
8/8, -6/8

1, -3/4

extract the factors
x=1
x-1=0
(x-1)

x= -3/4
4x= -3
4x+3=0
(4x+3)

(x-1)(4x+3)

except the way the equation is presented
the x's are on the other side of the factors
(1-x)(4+3x)

An expression cannot be solved. But in the case that the problem is this 4 - x - 3x^2 = 0
Just follow this:
4 - x -3x^2 = 0
Just send all the terms to the other side.
0 = 3x^2 + x - 4
Rearranging
3x^2 + x - 4 =0
Now you factorise
(3x + 4) (x - 1) = 0
Which gives you : x = -4/3 or x = 1

4 - x - 3x^2
is just the same as
- 3x^2 + 4 - x
or
- 3x^2 - x + 4
or
.....

there are 3 terms
-3x^2
-x
4

together they make the function f(x)
the order in wich the terms are written doesn't matter

you can look at iy like this

a + b + c = b + a + c = c + a + b = .....

if you think it is 'backwards'
then just reorganize the terms.

=
-3x^2 -x + 4

= (3x+4 )(-x+1 )

it makes no difference which order you write the terms.

1+2+3= 6

so does

3+2+1