you bring all xs on one side then you solve as quadratic analysis.


so you can simplify into

*Note that here 4 took the negative sign and 1 took the positive sign, as you want to numbers that if multiplied will give -4 and when added will give -3, and these are (-4 and 1)*
so now you can simply say that, for the equation to equal 0, x should be -1 or 4.
You can always substitute to make sure of your calculations.
Hope this helps.