Solve y=5/36*sqrt{1+4x}-1

$$y = \frac { 5 } { 36 } \cdot \sqrt { 1 + 4 x } - 1$$

Solve for x

$$x=\frac{324y^{2}}{25}+\frac{648y}{25}+\frac{1271}{100}$$

$$\frac{36y+36}{5}\geq 0$$

Steps for Solving Radical Equations

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

$$\frac{5}{36}\sqrt{1+4x}-1=y$$

Add 1 to both sides.

$$\frac{5}{36}\sqrt{1+4x}=y+1$$

Divide both sides of the equation by \frac{5}{36}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{5}{36}\sqrt{4x+1}}{\frac{5}{36}}=\frac{y+1}{\frac{5}{36}}$$

Dividing by \frac{5}{36} undoes the multiplication by \frac{5}{36}.

$$\sqrt{4x+1}=\frac{y+1}{\frac{5}{36}}$$

Divide y+1 by \frac{5}{36} by multiplying y+1 by the reciprocal of \frac{5}{36}.

$$\sqrt{4x+1}=\frac{36y+36}{5}$$

Square both sides of the equation.

$$4x+1=\frac{1296\left(y+1\right)^{2}}{25}$$

Subtract 1 from both sides of the equation.

$$4x+1-1=\frac{1296\left(y+1\right)^{2}}{25}-1$$

Subtracting 1 from itself leaves 0.

$$4x=\frac{1296\left(y+1\right)^{2}}{25}-1$$

Subtract 1 from \frac{1296\left(1+y\right)^{2}}{25}.

$$4x=\frac{1296y^{2}+2592y+1271}{25}$$

Divide both sides by 4.

$$\frac{4x}{4}=\frac{1296y^{2}+2592y+1271}{4\times 25}$$

Dividing by 4 undoes the multiplication by 4.

$$x=\frac{1296y^{2}+2592y+1271}{4\times 25}$$

Divide \frac{1271+2592y+1296y^{2}}{25} by 4.

$$x=\frac{324y^{2}}{25}+\frac{648y}{25}+\frac{1271}{100}$$

Solve for x (complex solution)

$$x=\frac{324y^{2}}{25}+\frac{648y}{25}+\frac{1271}{100}$$

$$y=-1\text{ or }arg(\frac{36y+36}{5})<\pi$$

Solve for y (complex solution)

$$y=\frac{5\sqrt{4x+1}}{36}-1$$

Solve for y

$$y=\frac{5\sqrt{4x+1}}{36}-1$$

$$x\geq -\frac{1}{4}$$