How do you solve \[2{{x}^{2}}-3x-7\] using quadratic formula?

Hint: We will solve this problem using the quadratic equation concept. As in the question it is given that we have to solve it using quadratic formula. We will substitute our values in the formula and perform the calculations that are needed to arrive at the solution.

Formula used:
Let the quadratic equation be \[a{{x}^{2}}+bx+c=0\]. Where a, b, c are the numerical coefficients given to solve the equation.
Now the quadratic formula derived to solve the equation and find the value of x is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]

Complete step-by-step answer:
 First, we have to know the quadratic equation form and its formula to solve the problem.
The standard quadratic form is
\[a{{x}^{2}}+bx+c\]. Where a, b, c are the numerical coefficients given to solve the equation.
Now the quadratic formula derived to solve the equation and find the value of x is
\[x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\]
Using this formula, we will get the roots of the given expression.
Given equation is
\[2{{x}^{2}}-3x-7\]
From the given equation we have to find the coefficients. They are
\[a=2\]
\[b=-3\]
\[c=-7\]
Now we have to substitute these values in the above formula to get the roots.
By substituting the values in the formula, we get
\[x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-\left( 4\times 2\times -7 \right)}}{2\times 2}\]
Now we have to simplify the above calculations to get the result.
To simplify we will start by simplifying the square root part in the expression.
\[\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}-\left( -56 \right)}}{2\times 2}\]
Now we can make sign of \[56\] as positive because we already know that
\[\left( - \right)\times \left( - \right)=\left( + \right)\]
So the equation will look like
\[\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{{{\left( -3 \right)}^{2}}+56}}{2\times 2}\]
Now we have to calculate \[{{\left( -3 \right)}^{2}}\] then the expression will be like
\[\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{9+56}}{2\times 2}\]
\[\Rightarrow x=\dfrac{-\left( -3 \right)\pm \sqrt{65}}{2\times 2}\]
As we already said two negative signs will give a positive then the expression will be like
\[\Rightarrow x=\dfrac{\left( 3 \right)\pm \sqrt{65}}{2\times 2}\]
\[\Rightarrow x=\dfrac{\left( 3 \right)\pm \sqrt{65}}{4}\]
Now we have to calculate the root of \[65\] then the expression will be like
\[\Rightarrow x=\dfrac{\left( 3 \right)\pm 8.0622577}{4}\]
Now we have to make it as two roots one by adding the two terms in numerator and another one by subtracting the two terms.
Then it will be like
  \[\Rightarrow x=\dfrac{\left( 3 \right)+8.0622577}{4}\]
\[\Rightarrow x=\dfrac{\left( 3 \right)-8.0622577}{4}\]
Now simplifying it we get
\[\Rightarrow x=\dfrac{11.0622577}{4}\]
\[\Rightarrow x=\dfrac{-5.0622577483}{4}\]
By dividing we will get x as
\[\Rightarrow x=2.765564437\]
\[\Rightarrow x=-1.26556443\]
So the x values we will get by solving the given equation are
\[x=2.765564437,x=-1.26556443\]

Note: we can solve this type of question in many ways by grouping the factor method and so on. But in the question it is clearly mentioned that we have to use quadratic formula so be specific while solving the equation. We can leave the roots in terms of roots of 65 also. We can substitute the value of x in the quadratic equation and cross check, but we must be cautious because depending on the decimals, the results will slightly vary.