Question 3471
1) x²-4=0
you bring the 4 to the other side: x²=4 thus x is 2 OR -2 because (-2)² gives also 4. 
2) x²+x=0
You have heard about  (this are the solutions of a second degree equation in the form of: ax²+bx+c=0) i guess. in this formula, a=1;b=1;c=0 so this gives:  and if you count a bit, you get that 0 and -1 are the roots. 
3)x^2+7x+12=0
Again the formula, and filled in, it gives: . The solutions are: -3 and -4 
4)x^2+5x=-4
Also here we can use the formula but we first have to put the -4 on the left side: x²+5x+4=0 (the polynome hase to be zero!) and then again you have to apply the formula. You can do it by now i guess? I find -1 and -4. 
5)3x^2+5x-10=0
THe same over and over again: a=3 b=5, c=-10. 
6)4x^2+3x+8=0
Again the same. This equation won't have any solutions because b²-4ac is smaller than 0 and in the plane of the real numbers that isn't possible. (in the complex plane it is!) 
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