How do you solve $2{{x}^{2}}-x-15=0$?

Hint: The equation $2{{x}^{2}}-x-15=0$, given in the above question is a quadratic equation. So we can use the quadratic formula which is given as $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ to solve it. For this, we have to substitute the values of the coefficients from the given equation into the quadratic formula. From the given equation, we have $a=2$, $b=-1$ and $c=-15$. On substituting these into the quadratic formula, we will obtain two solutions of the given equation.

Complete step by step solution:
The given equation is
$\Rightarrow 2{{x}^{2}}-x-15=0$
The above equation is in terms of the variable $x$ whose highest power is equal to two. This means that its degree is equal to two, or it is a quadratic equation. So we can solve it using the quadratic formula, which is given by
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
From the given equation, the values of the coefficients of ${{x}^{2}}$, $x$ and the constant terms can be respectively noted as
$\begin{align}
  & \Rightarrow a=2 \\
 & \Rightarrow b=-1 \\
 & \Rightarrow c=-15 \\
\end{align}$
Substituting these into the quadratic formula written above, we get
$\begin{align}
  & \Rightarrow x=\dfrac{-\left( -1 \right)\pm \sqrt{{{\left( -1 \right)}^{2}}-4\left( 2 \right)\left( -15 \right)}}{2\left( 2 \right)} \\
 & \Rightarrow x=\dfrac{1\pm \sqrt{1+120}}{4} \\
 & \Rightarrow x=\dfrac{1\pm \sqrt{121}}{4} \\
\end{align}$
We know that $\sqrt{121}=11$. Putting this above we get
$\begin{align}
  & \Rightarrow x=\dfrac{1\pm 11}{4} \\
 & \Rightarrow x=\dfrac{1+11}{4},x=\dfrac{1-11}{4} \\
 & \Rightarrow x=\dfrac{12}{4},x=\dfrac{-10}{4} \\
 & \Rightarrow x=3,x=-\dfrac{5}{2} \\
\end{align}$

Hence, the solutions of the given equation are $x=3$ and $x=-\dfrac{5}{2}$.

Note: We can also solve the given equation by factoring the LHS of the equation using the middle term splitting method. For this we have to split the middle term of $-x$ into the sum of two terms such that their product is equal to the product of the first term $\left( 2{{x}^{2}} \right)$ and the third term $\left( 15 \right)$, that is equal to $30{{x}^{2}}$. So we can split the middle term as $-x=5x-6x$ and take the common factors outside from each of the two pairs of the terms to factorize the LHS. Finally, on using the zero product rule, we will get the same solutions which we have obtained above.