Question 985339
You set each parenthetical term equal to zero and solve.
for the first, (3+x)=0, x= -3; the others are -7, and 2.
for the second, it is -8, +7, and the solution to x^2-2x+5=0.  That is done with the quadratic formula.
The result is x=(1/2) [2+/- sqrt (4-20)]=(1/2) [(2+/-sqrt (-16)]=1+/- 2i., because sqrt 16=4i and we take half.
;;
the third is 0,9,, and -(1/2)
the fourth is -4 and x^2-x+3 by the quadratic is (1/2) [1+/- sqrt (1-12)]= (1/2)+/- (1/2)sqrt (11 i)
;
The last: x=-1,3