Solve 3(-1/2x^2-x+3-x-3)/2=y

$$\frac{ 3(- \frac{ 1 }{ 2 } { x }^{ 2 } -x+3-x-3) }{ 2 } =y$$

Solve for y

$$y=-\frac{3x\left(x+4\right)}{4}$$

Solution Steps

Combine -x and -x to get -2x.

$$\frac{3\left(-\frac{1}{2}x^{2}-2x+3-3\right)}{2}=y$$

Subtract 3 from 3 to get 0.

$$\frac{3\left(-\frac{1}{2}x^{2}-2x\right)}{2}=y$$

Use the distributive property to multiply 3 by -\frac{1}{2}x^{2}-2x.

$$\frac{-\frac{3}{2}x^{2}-6x}{2}=y$$

Divide each term of -\frac{3}{2}x^{2}-6x by 2 to get -\frac{3}{4}x^{2}-3x.

$$-\frac{3}{4}x^{2}-3x=y$$

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

$$y=-\frac{3}{4}x^{2}-3x$$

Solve for x (complex solution)

$$x=-\frac{2\sqrt{9-3y}}{3}-2$$

$$x=\frac{2\sqrt{9-3y}}{3}-2$$

Solve for x

$$x=-\frac{2\sqrt{9-3y}}{3}-2$$

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