Solve y=x^2-lceil4rceil

$$y= { x }^{ 2 } - \lceil 4 \rceil$$

Solve for y

$$y=x^{2}-4$$

Solution Steps

The ceiling of a real number a is the smallest integer number greater than or equal to a. The ceiling of 4 is 4.

$$y=x^{2}-4$$

Solve for x (complex solution)

$$x=-\sqrt{y+4}$$

$$x=\sqrt{y+4}$$

Solve for x

$$x=\sqrt{y+4}$$

$$x=-\sqrt{y+4}\text{, }y\geq -4$$