Solve p_{1}:y=-1/4x^2+6

$$p _ { 1 } : y = - \frac { 1 } { 4 } x ^ { 2 } + 6$$

Solve for p_1

$$p_{1}=-\frac{y\left(x^{2}-24\right)}{4}$$

$$y\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4y, the least common multiple of y,4.

$$4p_{1}=-\frac{1}{4}x^{2}\times 4y+4y\times 6$$

Multiply -\frac{1}{4} and 4 to get -1.

$$4p_{1}=-x^{2}y+4y\times 6$$

Multiply 4 and 6 to get 24.

$$4p_{1}=-x^{2}y+24y$$

The equation is in standard form.

$$4p_{1}=24y-yx^{2}$$

Divide both sides by 4.

$$\frac{4p_{1}}{4}=\frac{y\left(24-x^{2}\right)}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$p_{1}=\frac{y\left(24-x^{2}\right)}{4}$$

Divide y\left(-x^{2}+24\right) by 4.

$$p_{1}=-\frac{yx^{2}}{4}+6y$$

Solve for x (complex solution)

$$x=-2\sqrt{-\frac{p_{1}}{y}+6}$$

$$x=2\sqrt{-\frac{p_{1}}{y}+6}\text{, }y\neq 0$$

Solve for x

$$x=2\sqrt{-\frac{p_{1}}{y}+6}$$

$$x=-2\sqrt{-\frac{p_{1}}{y}+6}\text{, }\left(y>0\text{ or }p_{1}\geq 6y\right)\text{ and }\left(y<0\text{ or }p_{1}\leq 6y\right)\text{ and }y\neq 0$$