Solution:
Given the equation:
[tex]2x^2-4x-3=x\text{ ----- equation 1}[/tex]
To solve for x,
step 1: Subtract x from both sides of the equation.
This gives
[tex]\begin{gathered} 2x^2-4x-x-3=x-x \\ \Rightarrow2x^2-5x-3=0\text{ ----- equation 2} \end{gathered}[/tex]
Step 2: From equation 2, solve for x by factorization.
Thus, we have
[tex]\begin{gathered} 2x^2-6x+x-3=0 \\ Group\text{ }the\text{ terms,} \\ (2x^2-6x)+(x-3)=0 \\ factor\text{ out the common terms,} \\ 2x(x-3)-1(x-3)=0 \\ \Rightarrow(2x-1)(x-3)=0 \\ thus, \\ 2x-1=0\text{ or x-3 =0} \\ when \\ 2x-1=0 \\ x=-\frac{1}{2} \\ when \\ x-3=0 \\ x=3 \end{gathered}[/tex]
Thus, the solution to the quadratic equation is
[tex]-\frac{1}{2}\text{ or 3}[/tex]
The correct option is
