63t2 - 38t - 5 = 0

Question

use the quadratiic formula

Answer 1

While Robert's answers are correct, he didn't find them by using the quadratic formula.

The formula is very difficult to write in this wordprocessor, but it goes like this:

     t  =  ( - b plus or minus the square root of ( b² - 4ac ) ) all over 2a

where    we look at the equation as   at²+bt+c=0.

So in your equation, a is 63, b is -38, and c is -5.

    so t = (- (-38)   plus or minus the square root of  (-38)² - 4*63*(-5)         all divided by 2*63.

    -(-38)  = 38                    (-38)² = 1444                    4*63*-5 = -1260                2*63 = 126

        38 = 38                        1444 - (-1260)   = 2704                          126 = 126

        38 = 38               square root of 2704 = 52                       126 = 126

so t is either     (38+52)/126                           or              (38-52)/126

which is either        5 / 7                               or                   -1 / 9.

Answer 2

Since 63*(-5) = 7(-45), and 7-45 = -38, the coefficient of t, you can factor it out as

 63t² - 38t - 5 = (9t + 1)(7t - 5) = 0

Answer: t = -1/9, 5/7