Solve 13-(3x-1)(3*+1)=2y-left(3x+2right)^2

$$13-(3x-1)(3 \times +1)=2y- { \left(3x+2 \right) }^{ 2 }$$

Solve for y

$$y=\frac{9x^{2}}{2}+\frac{3x}{2}+10$$

Steps for Solving Linear Equation

Multiply 3 and 1 to get 3.

$$13-\left(3x-1\right)\times 3=2y-\left(3x+2\right)^{2}$$

Use the distributive property to multiply 3x-1 by 3.

$$13-\left(9x-3\right)=2y-\left(3x+2\right)^{2}$$

To find the opposite of 9x-3, find the opposite of each term.

$$13-9x+3=2y-\left(3x+2\right)^{2}$$

Add 13 and 3 to get 16.

$$16-9x=2y-\left(3x+2\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+2\right)^{2}.

$$16-9x=2y-\left(9x^{2}+12x+4\right)$$

To find the opposite of 9x^{2}+12x+4, find the opposite of each term.

$$16-9x=2y-9x^{2}-12x-4$$

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

$$2y-9x^{2}-12x-4=16-9x$$

Add 9x^{2} to both sides.

$$2y-12x-4=16-9x+9x^{2}$$

Add 12x to both sides.

$$2y-4=16-9x+9x^{2}+12x$$

Combine -9x and 12x to get 3x.

$$2y-4=16+3x+9x^{2}$$

Add 4 to both sides.

$$2y=16+3x+9x^{2}+4$$

Add 16 and 4 to get 20.

$$2y=20+3x+9x^{2}$$

The equation is in standard form.

$$2y=9x^{2}+3x+20$$

Divide both sides by 2.

$$\frac{2y}{2}=\frac{9x^{2}+3x+20}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$y=\frac{9x^{2}+3x+20}{2}$$

Divide 20+3x+9x^{2} by 2.

$$y=\frac{9x^{2}}{2}+\frac{3x}{2}+10$$

Solve for x (complex solution)

$$x=\frac{\sqrt{8y-79}-1}{6}$$

$$x=\frac{-\sqrt{8y-79}-1}{6}$$

Solve for x

$$x=\frac{-\sqrt{8y-79}-1}{6}$$

$$x=\frac{\sqrt{8y-79}-1}{6}\text{, }y\geq \frac{79}{8}$$