Question

$6x {}^{2} + x - 1 = 0$
​

Answer 1

$6 {x}^{2} + x - 1 =0 \\ \\ Δ = {b}^{2} - 4 \times a \times c \\ Δ = {1}^{2} - 4 \times 6 \times ( - 1) \\ Δ = 1 + 24 \\ Δ = 25 \\ \\ x = \frac{ - 1± \sqrt{25} }{2 \times 6} = \frac{ - 1±5 }{12} \\ x1 = \frac{ - 1 + 5}{12} = \frac{1}{3} \\ x2 = \frac{ - 1 - 5}{12} = - \frac{ - 6}{12} = - \frac{ 1}{2}$

Answer 2

Equação do 2° grau

$6 {x}^{2} +x - 1 = 0 \\ \\ a = 6 \\ b = 1 \\ c = ( - 1) \\ \\ \Delta = {b}^{2} - 4ac \\ \Delta = {1}^{2} - 4 \cdot6 \cdot( - 1) \\\Delta = 1 - 24 \cdot( - 1) \\ \Delta = 1 + 24 \\ \Delta = 25 \\ \\ x = \frac{ - b\pm \sqrt{\Delta} }{2a} \\ \\ x = \frac{ - 1\pm \sqrt{25} }{2 \cdot6 } \\ \\ x = \frac{ - 1\pm 5 }{12} \\ \\ x_{1} = \frac{ - 1 + 5}{12} = \frac{ {4}^{\color{red} { \div 4}} }{ {12}^{\color{red} { \div 4}} } = \bf{ \frac{1}{3} } \\ \\ x_{2} = \frac{ - 1 - 5}{12} = - \frac{ {6}^{\color{red} { \div 6}} }{ {12}^{\color{red} { \div 6}} } = - \bf{ \frac{1}{2} }$

$\boxed{ \bf{S = \{ - \frac{1}{2} , \: \frac{1}{3} \} } }$