Are These Quadratic Solution Steps Correct?

[FAQ]

Just check these over and seen if there correct, thanks

#1
1-x=0.5x^2
Times everything by 2
2-2x=1x^2
Move it over
-1x^2 - 2x + 2=0
a=-1
b=-2
c=2

x=2 +/- root -2 - 4(-1)(2) all over -2

x=2 t/- root 6 all over -2

x=-1 +/- root -3

#2
2-x=x^2 + x
-x^2 -x -x + 2=0
a=-1
b=-2
c=2
same thing as #1 so if that correct then I guess this will be

#3
6(2-x)=3x^2 + 6x
I warn you, this could be wrong
12-6x=yada yada
-3x^2 -6x -6x + 12=0
a=-3
b=-6
c=12

x=6 +/- root -6 - 4(-3)(12) all over -6

x=6 +/- root 138 all over -6

x= -1 +/- root 23

Thanks for any help

 

[NateTG]

#3
6(2-x)=3x^2 + 6x

I think you've got some algebra errors in there:
$$6(2-x)=3x^2-6x$$
Divide both sides by $$3$$
$$2(2-x)=x^2-2x$$
multiply out
$$4-2x=x^2-2x$$
add $$2x-4$$ to both sides
$$0=x^2-4$$
$$x = \pm 2$$

 

[mathlete]

#3
6(2-x)=3x^2 + 6x
I warn you, this could be wrong
12-6x=yada yada
-3x^2 -6x -6x + 12=0
a=-3
b=-6
c=12

Since you have $$-6x-6x$$, $$b=-12$$

I think you've got some algebra errors in there:
$$6(2-x)=3x^2-6x$$
Divide both sides by $$3$$
$$2(2-x)=x^2-2x$$
multiply out
$$4-2x=x^2-2x$$
add $$2x-4$$ to both sides
$$0=x^2-4$$
$$x = \pm 2$$

I think the original equation was $$6(2-x)=3x^2+6x$$, not $$6(2-x)=3x^2-6x$$