How do you solve by completing the square \[2{x^2} - 7x - 15 = 0\]?

Hint: We will first divide the coefficients of each term by the coefficient of the leading term. Then, we will simplify the equation further and add a suitable constant to both sides to complete the square. Finally, we will take the square root on both sides and simplify it further to find the value of \[x\].

Formula used:
\[{(a - b)^2} = {a^2} - 2ab + {b^2}\]

Complete step by step solution:
The given equation is \[2{x^2} - 7x - 15 = 0\]. We have to solve this equation by completing the square.
First, let us divide the coefficient of every term by the coefficient of \[{x^2}\] i.e. 2. Therefore, we get
\[{x^2} - \dfrac{7}{2}x - \dfrac{{15}}{2} = 0\]
Adding \[\dfrac{{15}}{2}\] on both the sides, we get
\[ \Rightarrow {x^2} - \dfrac{7}{2}x = \dfrac{{15}}{2}\] ………\[\left( 1 \right)\]
Now, we will consider the coefficient of the \[x\] term i.e., \[ - \dfrac{7}{2}\]. We will take the half of this coefficient i.e.,
\[\dfrac{1}{2}\left( { - \dfrac{7}{2}} \right) = - \dfrac{7}{4}\]
We will now square this term and add the result to both sides of equation \[\left( 1 \right)\]. Therefore, we get
\[{x^2} - \dfrac{7}{2}x + {\left( { - \dfrac{7}{4}} \right)^2} = \dfrac{{15}}{2} + {\left( { - \dfrac{7}{4}} \right)^2}\]
\[ \Rightarrow {x^2} - \dfrac{7}{2}x + \left( {\dfrac{{49}}{{16}}} \right) = \dfrac{{15}}{2} + \left( {\dfrac{{49}}{{16}}} \right)\] ………\[\left( 2 \right)\]
We can see that equation \[\left( 2 \right)\] is of the form \[{(a - b)^2} = {a^2} - 2ab + {b^2}\], where \[a = x\] and \[b = \dfrac{7}{4}\]. Thus, we can write equation \[\left( 2 \right)\] as:
\[ \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} = \dfrac{{15}}{2} + \left( {\dfrac{{49}}{{16}}} \right)\]
Taking LCM in the RHS, we get
\[ \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} = \dfrac{{120}}{{16}} + \dfrac{{49}}{{16}}\]
Adding the terms, we get
\[ \Rightarrow {\left( {x - \dfrac{7}{4}} \right)^2} = \dfrac{{169}}{{16}}\]
Taking square root on both sides, we get
\[ \Rightarrow \sqrt {{{\left( {x - \dfrac{7}{4}} \right)}^2}} = \sqrt {\dfrac{{169}}{{16}}} \]
\[ \Rightarrow \left( {x - \dfrac{7}{4}} \right) = \pm \dfrac{{13}}{4}\]
Adding \[\dfrac{7}{4}\] to both the sides, we gt
\[ \Rightarrow x = \dfrac{7}{4} \pm \dfrac{{13}}{4}\]
Rewriting the equation, we get
\[ \Rightarrow x = \dfrac{{7 + 13}}{4}\] or \[x = \dfrac{{7 - 13}}{4}\]
Adding and subtracting the terms, we get
\[ \Rightarrow x = \dfrac{{20}}{4} = 5\] or \[x = - \dfrac{6}{4} = - \dfrac{3}{2}\]

Therefore the solution of the given equation is \[x = 5\] and \[x = \dfrac{{ - 3}}{2}\].

Note:
The given equation can also be solved by using the discriminant method.
For any quadratic equation \[a{x^2} + bx + c = 0\], the solution is \[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\], where the quantity \[{b^2} - 4ac\] is called the discriminant.
In the given equation, \[a = 2,b = - 7,c = - 15\]. So, we have the discriminant as
\[{b^2} - 4ac = {( - 7)^2} - 4 \times 2 \times ( - 15) = 49 + 120\]
\[ \Rightarrow {b^2} - 4ac = 169\]
Now, taking the square root of this discriminant, we have \[\sqrt {{b^2} - 4ac} = \sqrt {169} = 13\].
Hence, the solution is
\[x = \dfrac{{ - ( - 7) \pm 13}}{{2 \times 2}} = \dfrac{{7 \pm 13}}{4}\]
Therefore, the solutions are \[x = 5\] and \[x = - \dfrac{3}{2}\].