Question

2)y=5x^2+9x-2 1)При x=- 1/5 2)Нули функций

Answer 1

$y=5x^2+9x-2 \\ \\ 1)x=-1/5 \\ 5* \frac{1}{25} -9* \frac{1}{5}-2= \frac{1}{5}- \frac{9}{5}-2= \frac{1-9-10}{5}= -\frac{18}{5} =-3.6 \\ \\ 2)5x^2+9x-2=0 \\ D=81+40=121=11^2 \\ x_1=(-9-11)/10=-2 \\ x_2=(-9+11)/10=1/5$