Solve y=a(2x+1)^2+b

$$y = a ( 2 x + 1 ) ^ { 2 } + b$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=-\frac{b-y}{\left(2x+1\right)^{2}}\text{, }&x\neq -\frac{1}{2}\\a\in \mathrm{C}\text{, }&y=b\text{ and }x=-\frac{1}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$y=a\left(4x^{2}+4x+1\right)+b$$

Use the distributive property to multiply a by 4x^{2}+4x+1.

$$y=4ax^{2}+4ax+a+b$$

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

$$4ax^{2}+4ax+a+b=y$$

Subtract b from both sides.

$$4ax^{2}+4ax+a=y-b$$

Combine all terms containing a.

$$\left(4x^{2}+4x+1\right)a=y-b$$

Divide both sides by 4x^{2}+4x+1.

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

Dividing by 4x^{2}+4x+1 undoes the multiplication by 4x^{2}+4x+1.

$$a=\frac{y-b}{4x^{2}+4x+1}$$

Divide y-b by 4x^{2}+4x+1.

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

Solve for a

$$\left\{\begin{matrix}a=-\frac{b-y}{\left(2x+1\right)^{2}}\text{, }&x\neq -\frac{1}{2}\\a\in \mathrm{R}\text{, }&y=b\text{ and }x=-\frac{1}{2}\end{matrix}\right.$$

Steps for Solving Linear Equation

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

$$y=a\left(4x^{2}+4x+1\right)+b$$

Use the distributive property to multiply a by 4x^{2}+4x+1.

$$y=4ax^{2}+4ax+a+b$$

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

$$4ax^{2}+4ax+a+b=y$$

Subtract b from both sides.

$$4ax^{2}+4ax+a=y-b$$

Combine all terms containing a.

$$\left(4x^{2}+4x+1\right)a=y-b$$

Divide both sides by 4x^{2}+4x+1.

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

Dividing by 4x^{2}+4x+1 undoes the multiplication by 4x^{2}+4x+1.

$$a=\frac{y-b}{4x^{2}+4x+1}$$

Divide y-b by 4x^{2}+4x+1.

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

Solve for b

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