Question

Preciso de um "Help" 16x2-8x+1=0

Answer 1

$16 {x}^{2}-8x + 1 = 0$
a=16
b=-8
c=1


∆=48 - 4 × 16 × 1
∆=48 - 64
∆=16

$x = \frac{ - ( - 8) + - \sqrt{16} }{2 \times 16}$

$x = \frac{8 + - 4}{32}$
$x1 = \frac{8 + 4}{32 } = \frac{12}{32} \div 4 = \frac{3}{8}$
$x2 = \frac{8 - 4}{32 } = \frac{4}{32} \div 4 = \frac{1}{8}$