Are you wanting it factorised?
If so:
multiply the constant, in this case '-3' by the co-efficient of x^2, so:
(2x-3=6)
Take the 2 off the co-efficient and replace the constant, -3, with -6, so you have:
x^2 -5x -6
Find products of -6 that also add to make -5.
The products, in this case, are -6 and 1:
-6 multiplied by 1 = -6
-6 + 1 = -5
You now have both co-efficients of x, so remove the -5 in front of the x in the ORIGINAL equation: '2x^2-5x-3' and replace with '-6x + x' so it becomes:
2x^2-6x+x-3
Split the equation between the two co-efficients of x:
'2x^2-6x' and 'x-3'
Factorise these separately e.g.
'2x^2-6x' becomes 2x(x-3).
and,
'x-3' becomes 1(x-3).
As you can see the equation inside both brackets are the same so now take the numbers in front of both brackets:'2x' and '+1' and form a separate bracket: (2x+1)
Discard one of the (x-3) brackets and keep the other.
so in brackets: (2x+1) (x-3).
x=-1/2 or x=3.
Matt
2x^2-5x-3 show all the step please thanks
2 answers
In short:
2x^2-5x-3 becomes 2x^2-6x+x-3.
Split and factorise for:
2x(x-3) 1(x-3)
Which becomes:
(2x+1)(x-3)
x=-1/2 or x=3
2x^2-5x-3 becomes 2x^2-6x+x-3.
Split and factorise for:
2x(x-3) 1(x-3)
Which becomes:
(2x+1)(x-3)
x=-1/2 or x=3