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The quadratic formula is given by \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

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For the equation \(2x^2 - 4x - 6 = 0\) we identify \(a = 2\) \(b = -4\) \(c = -6\).

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First calculate the discriminant: \(\Delta = b^2 - 4ac = (-4)^2 - 4 \times 2 \times (-6) = 16 + 48 = 64\).

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\(x = \frac{-(-4) \pm \sqrt{64}}{2 \times 2} = \frac{4 \pm 8}{4}\).

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The solutions are \(x = \frac{4 + 8}{4} = 3\) and \(x = \frac{4 - 8}{4} = -1\).

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#### \(x = 3 -1\)

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The sum of the first \(n\) terms of an arithmetic sequence is given by \(S_n = \frac{n}{2}(2a + (n-1)d)\). If the first term \(a = 5\) the common difference \(d = 3\) and \(S_n = 155\) find \(n\).

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We substitute \(a = 5\) \(d = 3\) and \(S_n = 155\) into the formula:

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\(155 = \frac{n}{2}(2 \times 5 + (n-1) \times 3)\).

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