How do you solve using the quadratic formula $2{x^2} - 6 = - x$?

Hint: First move $x$ to the left side of the equation by adding $x$ to both sides of the equation. Next, compare the given quadratic equation to the standard quadratic equation and find the value of numbers $a$, $b$ and $c$ in the given equation. Then, substitute the values of $a$, $b$ and $c$ in the formula of discriminant and find the discriminant of the given equation. Finally, put the values of $a$, $b$ and $D$ in the roots of the quadratic equation formula and get the desired result.

Formula used:
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$

Complete step by step solution:
We know that an equation of the form $a{x^2} + bx + c = 0$, $a,b,c,x \in R$, is called a Real Quadratic Equation.
The numbers $a$, $b$ and $c$ are called the coefficients of the equation.
The quantity $D = {b^2} - 4ac$ is known as the discriminant of the equation $a{x^2} + bx + c = 0$ and its roots are given by
$x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$ or $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$
So, first move $x$ to the left side of the equation by adding $x$ to both sides of the equation.
$2{x^2} + x - 6 = 0$
Next, compare $2{x^2} + x - 6 = 0$ quadratic equation to standard quadratic equation and find the value of numbers $a$, $b$ and $c$.
Comparing $2{x^2} + x - 6 = 0$ with $a{x^2} + bx + c = 0$, we get
$a = 2$, $b = 1$ and $c = - 6$
Now, substitute the values of $a$, $b$ and $c$ in $D = {b^2} - 4ac$ and find the discriminant of the given equation.
$D = {\left( 1 \right)^2} - 4\left( 2 \right)\left( { - 6} \right)$
After simplifying the result, we get
$ \Rightarrow D = 1 + 48$
$ \Rightarrow D = 49$
Which means the given equation has real roots.
Now putting the values of $a$, $b$ and $D$ in $x = \dfrac{{ - b \pm \sqrt D }}{{2a}}$, we get
$x = \dfrac{{ - 1 \pm 7}}{{2 \times 2}}$
It can be written as
$ \Rightarrow x = \dfrac{{ - 1 \pm 7}}{4}$
$ \Rightarrow x = \dfrac{3}{2}$ and $x = - 2$
So, $x = \dfrac{3}{2}$ and $x = - 2$ are roots/solutions of equation $2{x^2} - 6 = - x$.

Therefore, the solutions to the quadratic equation $2{x^2} - 6 = - x$ are $x = \dfrac{3}{2}$ and $x = - 2$.

Note: We can check whether $x = \dfrac{3}{2}$ and $x = - 2$ are roots/solutions of equation $2{x^2} - 6 = - x$ by putting the value of $x$ in given equation.
Putting $x = \dfrac{3}{2}$ in LHS of equation $2{x^2} - 6 = - x$.
\[{\text{LHS}} = 2{\left( {\dfrac{3}{2}} \right)^2} - 6\]
On simplification, we get
\[ \Rightarrow {\text{LHS}} = \dfrac{9}{2} - 6\]
\[ \Rightarrow {\text{LHS}} = - \dfrac{3}{2}\]
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = \dfrac{3}{2}$ is a solution of equation $2{x^2} - 6 = - x$.
Putting $x = - 2$ in LHS of equation $2{x^2} - 6 = - x$.
\[{\text{LHS}} = 2{\left( { - 2} \right)^2} - 6\]
On simplification, we get
\[ \Rightarrow {\text{LHS}} = 8 - 6\]
\[ \Rightarrow {\text{LHS}} = 2\]
$\therefore {\text{LHS}} = {\text{RHS}}$
Thus, $x = - 2$ is a solution of equation $2{x^2} - 6 = - x$.
Final solution: Therefore, the solutions to the quadratic equation $2{x^2} - 6 = - x$ are $x = \dfrac{3}{2}$ and $x = - 2$.