$$ f(x)=\frac{3 x-5}{2 x^{2}+11 x-6} $$ will not be a real number when the denominator $$ \boldsymbol{T} $$ is zero. Part: $$ 1 / 3 $$ Part 2 of 3 That is, when $$ \begin{array}{l}2 x^{2}+11 x-6=0 \ (\square)(\square)=0\end{array} $$

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Solve the equation $$ 2x^2+11x-6=0 $$.
Solve the quadratic equation by following steps:
- step0: Solve by factoring:
  $$2x^{2}+11x-6=0$$
- step1: Factor the expression:
  $$\left(x+6ight)\left(2x-1ight)=0$$
- step2: Separate into possible cases:
  $$\begin{align}&2x-1=0\&x+6=0\end{align}$$
- step3: Solve the equation:
  $$\begin{align}&x=\frac{1}{2}\&x=-6\end{align}$$
- step4: Rewrite:
  $$x_{1}=-6,x_{2}=\frac{1}{2}$$
The solutions to the equation $$2x^2+11x-6=0$$ are $$x=-6$$ and $$x=\frac{1}{2}$$ or $$x=0.5$$.

Therefore, the denominator $$T$$ will be zero when $$x=-6$$ or $$x=\frac{1}{2}$$ or $$x=0.5$$.
The denominator $$T$$ is zero when $$x=-6$$ or $$x=0.5$$.

will not be a real number when the denominator is zero.
Part:
Part 2 of 3
That is, when

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will not be a real number when the denominator is zero. Part: Part 2 of 3 That is, when