How do you find the x intercepts for $3x^2 - 4x - 8 = 0$ ?

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Answer 1 · 2015-07-06T17:51:31

Solve: $y = 3x^2 - 4x - 8 = 0$

$D = d^2 = b^2 - 4ac = 16 + 96 = 112$ -> $d = +- 4sqrt7$

$x = 8/6 +- (4sqrt7)/6 = 4/3 +- (2sqrt7)/3$

How do you find the x intercepts for 3x^2 - 4x - 8 = 03x^2 - 4x - 8 = 0?