Question

16⋅z⋅(z+5)−(4⋅z+1)в квадрате=1

Answer 1

$16z(z+5)-(4z+1)^2=1 \\ 16z^2+80z-(16z^2+8z+1) = 1 \\ 16z^2+80z-16z^2-8z-1 = 1 \\ 80z-8z = 1 + 1 \\ 72z =2 \\ z = \frac{2}{72} \\ \\ z = \frac{1}{36}$