Solve ${x^2} + x - 6 = 0$

Hint: First we have to define what the terms we need to solve the problem are.
We can factorize the given polynomial and then we find the values of$x$.
The general quadratic polynomial form is $a{x^2} + bx + c = 0$

Complete step by step solution:
The given polynomial function is ${x^2} + x - 6 = 0$, we have to find the zeros of the polynomial (so that we can find$x$ ) and it is in a real value to variables.
Now we will factorize the polynomial by splitting the middle terms as we can write $x = 3x - 2x$
So, we get in middle term as ${x^2} + 3x - 2x - 6 = 0$ and simplifying this we get $({x^2} + 3x) - (2x + 6) = 0$
Now taking the common factors out, in first term $x$ is common and in second term $2$ is common thus on taking $x$ and $2$ common, we get $ x(x + 3) - 2(x + 3) = 0$
By separating the common multiples, we can write $ (x + 3) \times (x - 2) = 0$
Since it is in multiply so going zero to both the terms $ (x + 3) = 0$ or $(x - 2) = 0$
Now equating the values of $x$to zero then find the value of$x$,
$ \Rightarrow x = - 3$ Or $x = 2$
Therefore, the zeros of the polynomial ${x^2} + x - 6 = 0$ are $ - 3,2$.

Note: we can also use the quadratic formula $a{x^2} + bx + c = 0$ to find the polynomial $x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$ and it is known as discriminant formula
We need to put the values and simply the following
Let $a = 1,b = 1$ and $c = - 6$ now putting these know values in the quadratic formula we get
$x = \dfrac{{ - 1 \pm \sqrt {{1^2} - 4(1)( - 6)} }}{{2(1)}} = \dfrac{{ - 1 \pm 5}}{2}$ and simplifying the equation we get $ \Rightarrow x = - 3$ or $x = 2$
Hence, we get the same answer as above.