Solve y=-x^2-3x+1quadx_{0}

$$y = - x ^ { 2 } - 3 x + 1 \quad x _ { 0 }$$

Solve for x_0

$$x_{0}=x^{2}+3x+y$$

Solution Steps

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

$$-x^{2}-3x+1x_{0}=y$$

Add x^{2} to both sides.

$$-3x+1x_{0}=y+x^{2}$$

Add 3x to both sides.

$$1x_{0}=y+x^{2}+3x$$

Reorder the terms.

$$x_{0}=x^{2}+3x+y$$

Solve for x (complex solution)

$$x=\frac{-\sqrt{4x_{0}-4y+9}-3}{2}$$

$$x=\frac{\sqrt{4x_{0}-4y+9}-3}{2}$$

Solve for x

$$x=\frac{-\sqrt{4x_{0}-4y+9}-3}{2}$$

$$x=\frac{\sqrt{4x_{0}-4y+9}-3}{2}\text{, }y\leq x_{0}+\frac{9}{4}$$