Помогите пожалуйста решить..

Решите задачу:

$1) 3y ^{2} -y+2\\y= \frac{1}{2} =0.5 \\ 3*0.5 ^{2} -0.5+2=3*0.25-0.5+2=0.75-0.5+2=0.25+2=\\=2.25 \\ \\ 2)0.6 x^{3} +0.4 x^{2} +3 \\ x=-2 \\ 0.6*(-2) ^{3} +0.4*(-2) ^{2} +3=0.6*(-8)+0.4*4+3= \\ =-4.8+1.6+3=-0.2 \\ \\ 3) \frac{b+d}{c} \\ b=6.3;d=-0.9;c=2.5 \\ \frac{6.3-0.9}{2.5} = \frac{5.4}{2.5} =2.16$

$4) \sqrt{-8x+y ^{2} } \\ x=-154 \\ \sqrt{-8*(-154)+y ^{2} } = \sqrt{1232+y ^{2} } \\ \\ 5) \sqrt{b ^{2} +b ^{2} } \\ a=28;b=-96 \\ \sqrt{2 8^{2} +(-96) ^{2} } = \sqrt{784+9216} = \sqrt{10000} =100$