How do you solve $2x^2 - 28x + 98 = 0$ using the quadratic formula?

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Answer 1 · 2016-03-19T04:53:20

Double root at x = 7

$y = x^2 - 14x + 49 = 0$
$D = d^2 = b^2 - 4ac = 196 - 196 = 0$
There is a double-root at $x = -b/(2a) = 14/2 = 7$

How do you solve 2x^2 - 28x + 98 = 02x^2 - 28x + 98 = 0 using the quadratic formula?