Question

решите пж через дискрименант
2(n-4)^2+(n^2-3n-2n+6)=62

Answer 1

2*(n^2-8n+16)+n^2-5n+6=62
2n^2-16n+32+n^2-5n+6-62=0
3n^2-21n-24=0
n^2-7n-8=0
D=(-7)^2-4*1*(-8)=49+32=81
n1=$frac{7+9}{2}=8$
n2=$frac{7-9}{2}=-1$