Show your solution need an answer, please need an answer, please.

As follows,

Step-by-step explanation:

You can solve these either by factoring or using the quadratic formula or other ways,

1. x^2+2x+1=0

Factoring,

(x+1)^2=0

x+1=0

x=-1

2.y^2-y-2=0

Using quadratic formula,

(-b±√(b^2-4ac))/2a

(--1±√(-1^2-4*1*-2))/2*1

(1±√(1+8))/2

(1±√(9))/2

(1±3)/2

y=(1±3)/2

y=(1+3)/2    y=(1-3)/2  

y=4/2          y=-2/2

y=2              y=-1

3.y^2+y+1=0

Using quadratic formula,

(-b±√(b^2-4ac))/2a

(-1±√(1^2-4*1*1))/2*1

(-1±√(1-4))/2

(-1±√(3))/2

No real roots

4. x^2+9x+18=0

Using quadratic formula,

(-9±√(9^2-4*1*18))/2*1

(-9±√(81-72))/2

(-9±√(9))/2

(-9±3)/2

x=(-9+3)/2  x= (-9-3)/2

x=-6/2         x=-12/2

x=-3           x=-6

5.x^2+5x+6=0

Using quadratic formula,

(-5±√(b^2-4ac))/2a

(-5±√(5^2-4*1*6))/2*1

(-5±√(25-24)/2

(-5±1)/2

x=(-5+1)/2  x=(-5-1)/2

x=-4/2       x=-6/2

x=-2         x=-3

6. 3n^2-13n-10=0

Using quadratic formula,

(-b±√(b^2-4ac))/2a

(-(-13)±√(-13^2-4*3*(-10)))/2*3

(13±√(169-(-120))/6

(13±√289)/6

The discriminant b^2−4ac>0

So there are 2 real roots.

x=(13±17)/6

x=30/6 or x=-4/6

x=5 or x=-2/3

i would have to say it is c.