Question

${3x}^{2} - 16x + 5 = 0$equação de 2° grau!

Answer 1

$Δ = {b}^{2} - 4.a.c$
${x}^{1} \: e \: \: {x}^{2} = \frac{ - b \: \: \: + - \sqrt{delta} }{2 \times a}$
calculando Δ:


$Δ = {( - 16)}^{2} - 4 \times 3 \times 5$
$Δ = 256 - 12 \times 5$
$Δ = 256 - 60 \\ Δ = 196$

${x}^{1} = \: \frac{ - ( - 16) + \sqrt{196} }{2 \times 3}$
${x}^{1} = \frac{16 + 14}{6} \\ {x}^{1} = \frac{30}{6} \\ {x}^{1} = 5$

${ x}^{2} = \frac{ - ( - 16) - \sqrt{196} }{2 \times 3}$
${x}^{2} = \frac{16 - 14}{6}$
${x}^{2} = \frac{2}{6} \\ {x}^{2} = \frac{1}{3}$