How to factor $2x^2-x-3$?
Here's a trick: multiply the first and last coefficients and see which factor pairs add to the middle coefficient.
In this case, $2\times (-3) = -6$. The factor pairs of $-6$ are $1,-6$, $2,-3$, $-1,6$, and $-2,3$. Since $2+(-3) = -1$, we can break up the middle coefficient and factor by grouping: \begin{align*} 2x^2 - x - 3 &= 2x^2 + (2-3)x - 3 \\ &= 2x^2 + 2x - 3x - 3 \\ &= 2x(x + 1) - 3(x + 1) \\ &= (2x - 3)(x+1). \end{align*}
The rule comes from the distributive rule: $(ax + b)(cx + d) = (ac)x^2 + (ad + bc)x + bd$. The first coefficient is $ac$ and the last is $bd$, they multiply together to $abcd$, and the middle coefficient is $ad + bc$, the sum of a factor pair of $abcd$.
Another way to do it is to use the quadratic formula to find the roots of the polynomial: $a = 2$, $b=-1$, and $c = -3$, so $$\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} = \frac{1 \pm\sqrt{1 - -24}}{4} = \frac{1\pm 5}{4} = \frac{3}{2}, -1,$$ which gives a factorization into the monic polynomials $x+1$ and $2x - 3$.
Note that $2-3=-1,$ so that $$\begin{align}2x^2-x-3 &= 2x^2+(2-3)x-3\\ &= 2x^2+2x-3x-3\\ &= 2x(x+1)-3(x+1)\\ &= (x+1)(2x-3).\end{align}$$
You can use several methods to solve the equation. Firstly you can use the quadratic formula
$$\frac{-b \pm\sqrt{b^2 - 4ac}}{2a}$$
or you can use decomposition.
Let's try decomposition first. Look for two numbers that multiple to $-6$ and add to $-1$. These numbers would be $+2$ and $-3$. So, we would place those for $-x$ and have the following:
$$2x^2-3x+2x-3$$
Now here you can factor out $x$ from the first two terms.
$$x(2x-3)+(2x-3)$$
Notice the coefficient on $2x-3$ is $1$. So this means that our equation is $x+1$ and $2x-3$. Therefore, giving us the solution
$$ (2x-3) = 0 \, \text{or} \, (x+1) = 0$$
Now, if you want to use the quadratic formula then you can do this. $a = 2$, $b = -1$ and $c = -3$. So we will have:
$$\frac{6 \pm\sqrt{(-1)^2 - 4(2)(-3)}}{2(2)}$$
When you apply this formula twice, you will get the roots which are $x = \frac{3}{2}$ and $x = -1$. If you have any questions, let me know. I will try to clarify as much as possible.