How do you solve with working out 3x(2x-3)-5(2x-3)?

1.Multiply whatever is inside the brackets
(2x-3)=-6
2. Multiply the two terms on either side of '-'
3x(-6)=-18
5x(-6)=-30
3. Since the second term has a negative sign and already, we get -(-30)=+30
4. Add the two terms:
-18+30=12

x^3 - 5 = 3 ==> x^3 - 8 = 0 ==> (x - 2)(x^2 + 2x + 4) = 0 ==> x - 2 = 0 ==> x = 2 ==> x^2 + 2x + 4 = 0 A = a million B = 2 C = 4 x = [-B +/- ?(B^2 - 4AC)] / 2A ==> x = [-2 +/- ?(2^2 - 4*a million*4)] / (2*a million) ==> x = [-2 +/- ?(4 - sixteen)] / 2 ==> x = (-2 +/- ?-12) / 2 ==> x = (-2 +/- ?-a million?4?3) / 2 ==> x = (-2 +/- 2i?3) / 2 ==> x = -a million +/- i?3 recommendations: x = 2 x = -a million + i?3 x = -a million - i?3 the different 2 i would be unable to appreciate what the equations are.

you can't solve it, but you can simplify it. Because this simplify question

simplify
(3x-5)(2x-3)

solve
(3x-5)(2x-3) = 0

x= 5/3 and 3/2

Your Question is to solve 3x(2x-3)-5(2x-3)
I am assuming x is multiplication and not x as "ex"
solve within the brackets 3x(-6)-5(-6)
then solve the multiplication part -18+30
--> [because -5x-6=+30]
then do the addition/substraction part +12
the answer is +12.

The word "SOLVE" is applied to equations and these require an = sign.
What we have here is an expression that has to be simplified.

( 2x - 3 ) ( 3x - 5 )

OR

6x^2 - 19 x + 15

not positive but think:
Distribute the outside terms and get
(6x^2-9x) - (10x-15)
then combine like terms
6x^2 - 19x -15

hope im right...

3x(2x - 3) - 5(2x - 3)
= (2x - 3)(3x - 5)