Can someone check my work? (Algebra)?

The easiest way to do this is to let x = z/2 + 1

Now you have 2x^2 - 3x = -1. Then add 1 to both sides

2x^2 - 3x + 1 = 0

Then factor by FOIL: (2x-1)(x-1) = 0

so either 2x-1 = 0 meaning x = 1/2, or x-1 = 0 meaning x=1

Then put back z/2 + 1 for x:

z/2 + 1 = 1/2 or z/2 + 1 = 1

Solve these for z.

To solve for z, you must find the value of z.

2[z/2+1]^2=2[z^2/4+z+1]=z^2/2+2z+2 --- (1)

-3[z/2+1]=-3z/2-3 -----(2)

we have (1)+(2)=-1

z^2/2+2z+2-3z/2-3+1=0

z^2+4z-3z=0

z^2+z=0

z(z+1)=0

z=0 or z=-1

0=2[z/2+1]^2-3[z/2+1]+1

let a=[z/2+1], then you'll have a quadratic equation: 0=2a^2 - 3a+1

solving for a using the quadratic formula will yield a=1 or 1/2

then, substituting it back and solving for z will give you z=0 or -1.

=)