Solve Delta=y^2-4*L-4

$$\Delta = y ^ { 2 } - 4 \cdot L - 4$$

Solve for L

$$L=\frac{y^{2}-\Delta -4}{4}$$

Steps for Solving Linear Equation

Swap sides so that all variable terms are on the left hand side.

$$y^{2}-4L-4=\Delta$$

Subtract y^{2} from both sides.

$$-4L-4=\Delta -y^{2}$$

Add 4 to both sides.

$$-4L=\Delta -y^{2}+4$$

The equation is in standard form.

$$-4L=4+\Delta -y^{2}$$

Divide both sides by -4.

$$\frac{-4L}{-4}=\frac{4+\Delta -y^{2}}{-4}$$

Dividing by -4 undoes the multiplication by -4.

$$L=\frac{4+\Delta -y^{2}}{-4}$$

Divide \Delta -y^{2}+4 by -4.

$$L=\frac{y^{2}}{4}-\frac{\Delta }{4}-1$$

Solve for y (complex solution)

$$y=-\sqrt{4L+\Delta +4}$$

$$y=\sqrt{4L+\Delta +4}$$

Solve for y

$$y=\sqrt{4L+\Delta +4}$$

$$y=-\sqrt{4L+\Delta +4}\text{, }\Delta \geq -\left(4L+4\right)$$