I have two questions:
1. Photomath tells me the first inequality is impossible because the left side is always bigger than zero. First of all, I thought the way to solve inequalities of second degree was to use the specific formula, not by just looking at it. If you actually use the formula on inequality nr20 it yelds a numeric result, so I don't get it: when do we use the formula?
2. Assuming nr20 was impossible, then shouldn't the other exercise also yeld the same result? The left side will never be less, nor equal to zero, because since it has a positive exponent whatever sign the x has it's always going to be turned into a + sign and therefore get added to the 3, that also has a + sign. Since both are plus and 3 is not zero, how come is the result x=-3 and not impossible?
Thank you for your attention