formulas for the roots x1, x2, x3 of cubic equations ax^3 +bx^2 + cx +d =0 where a != 0? 2. Viéte solved the quadratic equation x^2+ax = b by substituting x = y -a/2 . This produces a quadratic in y in which the first-degree term is missing. Use Viéte's method to solve the quadratic equations: \text { (a) } x^{2}+10 x=144 \text { (b) } x^{2}+12 x=64
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