In fact, the simple answer is YES. It is trivial. No loop required.
You have a set of QUADRATIC EQUATIONS!!!!!! Surely you learned the quadratic formula for the roots of a quadratic equation?
tic
N = 1e6;
coef = rand(N,3);
D = sqrt(coef(:,2).^2 - 4*coef(:,1).*coef(:,3));
R = [(-coef(:,2) + D)./(2*coef(:,1)), (-coef(:,2) - D)./(2*coef(:,1))];
toc
Elapsed time is 0.115124 seconds.
size(R)
ans =
1000000 2
One million sets of roots, computed in around 1/10 of a second. Since my coefficients are randomly generated, many of those roots will be complex. As a test to verify it worked...
coef(1,:)
ans =
0.562209637790579 0.0918349300613903 0.952524012642822
R(1,:)
ans =
-0.0816732086115523 + 1.29906892840699i -0.0816732086115523 - 1.29906892840699i
roots(coef(1,:))
ans =
-0.0816732086115523 + 1.29906892840699i
-0.0816732086115523 - 1.29906892840699i
Hopefully your quadratics will be better behaved, and have real roots.
