Question

16⋅z⋅(z+5)−(4⋅z+1)в квадрате=1

Answer 1

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