I have to code this in Matlab but a written out version work

Program 5-Roots (homework) CSCI 251 For program 5, we will revisit program 2, and rewrite it using a function and a main program instead. The two roots of a quadratic equation ax2 +bx +c can be obtained using the following formula: -b+ b2-4ac -b-vb2-4ac 2a 2a b2- 4ac is called the discriminant of the quadratic equation, and its value determines the number of roots as follows: b2-4ac >0, two real roots: rl and r2 b2-4ac-0, one real root:r b2-4ac

Solution

roots_yourLastName.m

function [numOfRoots] = roots_yourLastName( a, b, c)
    discriminant = b*b - 4*a*c;
    if discriminant < 0
        numOfRoots = 0;
    elseif discriminant == 0
        numOfRoots = 1;
    else
        numOfRoots = 2;
    end
end

rootsTest_yourLastName.m

function [] = rootsTest_yourLastName()
    a = input(\'Enter the A coefficient: \');
    b = input(\'Enter the B coefficient: \');
    c = input(\'Enter the C coefficient: \');
    numOfRoots = roots_yourLastName(a,b,c);
    fprintf(\'%d, %d, and %d coefficients have\',a,b,c);
    discriminant = b*b - 4*a*c;
    if numOfRoots==0
        fprintf(\' no real roots\ \');
    elseif numOfRoots==1
        fprintf(\' one real root:\ \');
        fprintf(\'\\tr=%.2f\ \', (-b*1.0)/2*a );
    else
        fprintf(\' two real roots:\ \');
        fprintf(\'\\tr1 = %.2f\ \',(-b+sqrt(discriminant)+0.0)/2*a);
        fprintf(\'\\tr2 = %.2f\ \',(-b-sqrt(discriminant)+0.0)/2*a);
    end
end