Solve the following equations: 3x2 − 5 = 3 and 22x+2 - 2^x+1 + 1 = 2^x

Add 5 to both sides of the equation:

$$3x^2 = 3 + 5$$ $$3x^2 = 8$$

Remember to include both positive and negative roots.

$$x = \pm\sqrt{\frac{8}{3}}$$ $$x = \pm\frac{\sqrt{8}}{\sqrt{3}}$$ $$x = \pm\frac{2\sqrt{2}}{\sqrt{3}}$$

Multiply the numerator and denominator by $\sqrt{3}$:

$$x = \pm\frac{2\sqrt{2} \cdot \sqrt{3}}{\sqrt{3} \cdot \sqrt{3}}$$ $$x = \pm\frac{2\sqrt{6}}{3}$$

The solutions for the first equation are:

$$\boxed{x=\pm\frac{2\sqrt{6}}{3}}$$

For the second equation: $2^{2x+2} - 2^{x+1} + 1 = 2^x$

$$2^{2x+2} = 2^{2x} \cdot 2^2 = (2^x)^2 \cdot 4$$ $$2^{x+1} = 2^x \cdot 2^1$$

Substitute these into the equation:

$$4 \cdot (2^x)^2 - 2 \cdot (2^x) + 1 = 2^x$$

Substitute $y$ into the equation. Note that for real $x$, $y$ must be a positive real number ($y > 0$).

$$4y^2 - 2y + 1 = y$$

$$4y^2 - 2y - y + 1 = 0$$ $$4y^2 - 3y + 1 = 0$$

Here, $a=4$, $b=-3$, $c=1$.
First, calculate the discriminant $\Delta = b^2 - 4ac$:

$$\Delta = (-3)^2 - 4(4)(1)$$ $$\Delta = 9 - 16$$ $$\Delta = -7$$

Since the discriminant $\Delta$ is negative ($\Delta < 0$), there are no real solutions for $y$.
Because $y = 2^x$ must be a positive real number, and there are no real values of $y$ that satisfy the quadratic equation, there are no real solutions for $x$.

The second equation has:

$$\boxed{\text{No real solutions for } x}$$