假设，请用求根公式计算，并总结两根和与系数a,b,c的关系？

假设，ax²+bx+c根的判别式=b²-4ac≥0，请用求根公式计算，x₁*x₂和x₁+x₂ax²+bx+c=0x₁=[-b+√(b²-4ac)]/2ax₂=[-b-√(b²-4ac)]/2ax₁+x₂=[-b+√(b²-4ac)]/2a+[-b-√(b²-4ac)]/2a =[-b+√(b²-4ac)-b-√(b²-4ac)]/2a =-2b/2a =-b/ax₁·x₂=[-b+√(b²-4ac)]/2a·[-b-√(b²-4ac)]/2a =-[√(b²-4ac)-b][√(b²-4ac)+b]/4a² =-[√(b²-4ac)²-b²]/4a² =-(b²-4ac-b²)/4a² =4ac/4a² =c/a总结两根x₁和x₂与系数a,b,c的关系x₁+x₂=-b/ax₁·x₂=c/a