## C Program To Find Roots of Quadratic Equation

Learn How To Find Roots of Quadratic Equation in C Programming Language using **If – Else** Block Structure. This Code To Calculate Quadratic Roots of an Equation is without Functions and also has an Output Screen displayed at the bottom of this page.

#### What is a Quadratic Equation?

The Standard Form of a Quadratic Equation is **ax ^{2} + bx + c = 0,** where x is a variable entity and a, b, c are constant values which cannot be changed.

#### Formula For Quadratic Equation

**z = −b ± √(b ^{2 }− 4ac) / 2a**

#### Conditions For Discriminants

- If
**b**is^{2 }− 4ac**Negative**, Two Complex Solutions are possible - If
**b**is^{2 }− 4ac**Positive**, Two Real Solutions are possible - If
**b**, One Real Solution is possible^{2 }− 4ac = 0

**Also Read: C Program To Remove Vowels From A String**

#### C Program To Find Roots of Quadratic Equation

#include<stdio.h> #include<math.h> int main() { float root1, root2, discriminant; float a, b, c; printf("\nEnter The Values\n"); printf("\nA:\t"); scanf("%f", &a); printf("\nB:\t"); scanf("%f", &b); printf("\nC:\t"); scanf("%f", &c); discriminant = b*b - 4*a*c; if(discriminant < 0) { printf("\nRoots Are Imaginary\n"); } else { root1 = (-b + sqrt(discriminant)) / (2*a); root2 = (-b - sqrt(discriminant)) / (2*a); printf("\nRoot 1 = %f\n", root1); printf("\nRoot 2 = %f\n", root2); } printf("\n"); return 0; }

**Note:** If you Compile this Code in Linux, you will get a **Linker Error**. It is because the implementation of **sqrt() method** is missing and it is defined in** libm **library in GNU GCC Compiler. Therefore, you have to explicitly apply it while compilation.

**Command: gcc test.c -lm**

**Also Read: Find Sum of Lower Triangular Elements of Matrix C Program**

#### Output

If you have any compilation errors or doubts in this Program to Find Roots of Quadratic Equation in C Programming Language, let us know about in the Comment Section below.

It is really a short and simple code for finding quadratic roots in c language. Thanks.