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2014-06-29T12:01:26-03:00
1) 2x^{2}  -  \frac{19}{15}x + \frac{2}{5}  = 0

Δ=b²-4ac

Δ= ( \frac{19}{15} )^{2}  -4.2. \frac{2}{5}

Δ= \frac{361}{225} - \frac{16}{5}

Δ= \frac{361-720}{225}

Δ=- \frac{-359}{225}

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