Given:
The equation is:
[tex](2x+1)(x-2)=0[/tex]
To find:
The values of a, b and b from the standard form.
Solution:
The standard form of a quadratic equation is:
[tex]ax^2+bx+c=0[/tex] ....(i)
We have,
[tex](2x+1)(x-2)=0[/tex]
In can be rewritten as
[tex]2x(x)+(2x)(-2)+(1)(x)+(1)(-2)=0[/tex]
[tex]2x^2-4x+x-2=0[/tex]
[tex]2x^2-3x-2=0[/tex] ...(ii)
On comparing (i) and (ii), we get
[tex]a=2,b=-3,c=-2[/tex]
Therefore, the correct option is C.
