x²+3x-5=a(x+1)x²+(3-a)x-(a+5)=0△=(3-a)²+4(a+5)=a²-2a+29>0X=[-(3-a)土√(a²-2a+29)]/2∴X1=[a-3+√(a²-2a+29)]/2,X1=[a-3-√(a²-2a+29)]/2