To complete the problem, nestedifs is use where the first if is dedicated to variable a while the second if is for the variable d. Thus, now the root when a=0 is also being computed. d = b^2-4*a*c; if a-=0 ! if (d==0) ! When a=0, the quadratic equation becomes linear equation. Thus, only one root available and -b/ (2*a); х2 %3D х1; х1 elseif (d>0) computed as x1 in the (-b+sqrt(d))/ (2*a); (-b-sqrt(d)) 7 (2* a); х1 outer else. x2 %3D else disp('complex root') end else x1 = -c/b; end

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