Strictly speaking, the quadratic formula is nor required in this instance because the trinomial 10x^2-11x-84 is factorable. FYI, any quadratic trinomial, ax^2+bx+c, is factorable whenever its discriminant, b^2-4ac, is a perfect square. In the present case, b^2-4ac = (-11)^2 - 4(10)(-84) = 3481, which is indeed a perfect square (Any integer, k, is a perfect square if sqrt(k) is itself an integer. Because sqrt(3481)=59, and 59 is an integer, it follows that 3481 is by definition a perfect square). Nevertheless, were this my problem, I would use the q.formula just the same, as the coefficients 10 and 84 make for more factor combinations than I care to test.