Question

6 x² + 3x = 1 + 2 x

qual o delta? e o baskara?​

Answer 1

$6 {x}^{2} + 3x = 1 + 2x \\ 6x^{2} + 3x - 1 - 2x = 0 \\ \\ 6 {x}^{2} + x - 1 = 0 \\ Δ = b^{2} - 4.a.c \\ Δ = 1^{2} - 4.6.( - 1) \\ Δ = 1 + 24 = 25 \\ \\ {x}^{2} = \frac{ - b± \sqrt{Δ} }{2.a} \\ \\ {x}^{2} = \frac{ - 1± \sqrt{25} }{2.6} \\ \\ {x}^{1} = \frac{ - 1 + 5}{12} = \frac{4}{12} \div 2 = \frac{2}{6} = \frac{1}{3} \\ \\ {x}^{2} = \frac{ - 1 - 5}{12} = \frac{ - 6}{12} \div 2 = \frac{ - 3}{6} \div 3 = \frac{ - 1}{2}$

Δ = 25

x¹ = 1/3

x² = -1/2