Question

(4-у)²-у(у-6) при у= - 1/20

Answer 1

$\displaystyle (4-y)^2-y(y-6)=16-8y+y^2-y^2+6y=16-8y+6y=16-2y\\\\\\y=-\frac{1}{20} \\\\\\16-2*(-\frac{1}{20} )=16+2*\frac{1}{20} =16+\frac{1}{20} =\frac{160+1}{10}=\frac{161}{10} \;\;\;\;ili\;\;\;16.1$