c语言程序设计题目:计算一元二次方程的根【程序填空】#include//elseif(【?】){x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);printf(quot;hasdistinctrealroots:%8.4fand%.4fnquot;,x1,x2);}else{realpart=-b/(2*a);imagpart=sqrt(-di
<p>问题：c语言程序设计题目:计算一元二次方程的根【程序填空】#include//elseif(【?】){x1=(-b+sqrt(disc))/(2*a);x2=(-b-sqrt(disc))/(2*a);printf(quot;hasdistinctrealroots:%8.4fand%.4fnquot;,x1,x2);}else{realpart=-b/(2*a);imagpart=sqrt(-di<p>答案：↓↓↓<p class="nav-title mt10" style="border-top:1px solid #ccc;padding-top: 10px;">刘士兴的回答：<div class="content-b">网友采纳 #include // elseif(disc>0) { x1=(-b+sqrt(disc))/(2*a); x2=(-b-sqrt(disc))/(2*a); printf("hasdistinctrealroots:%8.4fand%.4fn",x1,x2); } else { realpart=-b/(2*a); imagpart=sqrt(-disc)/(2*a); printf("hascomplexroots:n"); printf("%8.4f+%.4fin",realpart,imagpart); printf("%8.4f-%.4fin",realpart,imagpart); } } } 2-3-5 theequationhasdistinctrealroots:2.5000and-1.0000 Pressanykeytocontinue
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