Solve (x+b/2a)^2=b^2/4a^2-4ac/4a^2

$$( x + \frac { b } { 2 a } ) ^ { 2 } = \frac { b ^ { 2 } } { 4 a ^ { 2 } } - \frac { 4 a c } { 4 a ^ { 2 } }$$

Solve for b (complex solution)

$$\left\{\begin{matrix}b=-ax-\frac{c}{x}\text{, }&x\neq 0\text{ and }a\neq 0\\b\in \mathrm{C}\text{, }&c=0\text{ and }x=0\text{ and }a\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4a^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2a}{2a}.

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

Since \frac{x\times 2a}{2a} and \frac{b}{2a} have the same denominator, add them by adding their numerators.

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

To raise \frac{x\times 2a+b}{2a} to a power, raise both numerator and denominator to the power and then divide.

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

Express 4\times \frac{\left(x\times 2a+b\right)^{2}}{\left(2a\right)^{2}} as a single fraction.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x\times 2a+b\right)^{2}.

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

Expand \left(2a\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Cancel out 4 in both numerator and denominator.

$$\frac{4a^{2}x^{2}+4abx+b^{2}}{a^{2}}a^{2}=b^{2}-4ac$$

Express \frac{4a^{2}x^{2}+4abx+b^{2}}{a^{2}}a^{2} as a single fraction.

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

Cancel out a^{2} in both numerator and denominator.

$$4a^{2}x^{2}+4abx+b^{2}=b^{2}-4ac$$

Subtract b^{2} from both sides.

$$4a^{2}x^{2}+4abx+b^{2}-b^{2}=-4ac$$

Combine b^{2} and -b^{2} to get 0.

$$4a^{2}x^{2}+4abx=-4ac$$

Subtract 4a^{2}x^{2} from both sides.

$$4abx=-4ac-4a^{2}x^{2}$$

The equation is in standard form.

$$4axb=-4a^{2}x^{2}-4ac$$

Divide both sides by 4ax.

$$\frac{4axb}{4ax}=-\frac{4a\left(ax^{2}+c\right)}{4ax}$$

Dividing by 4ax undoes the multiplication by 4ax.

$$b=-\frac{4a\left(ax^{2}+c\right)}{4ax}$$

Divide -4a\left(c+ax^{2}\right) by 4ax.

$$b=-ax-\frac{c}{x}$$

Solve for b

$$\left\{\begin{matrix}b=-ax-\frac{c}{x}\text{, }&x\neq 0\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&c=0\text{ and }x=0\text{ and }a\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 4a^{2}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply x times \frac{2a}{2a}.

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

Since \frac{x\times 2a}{2a} and \frac{b}{2a} have the same denominator, add them by adding their numerators.

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

To raise \frac{x\times 2a+b}{2a} to a power, raise both numerator and denominator to the power and then divide.

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

Express 4\times \frac{\left(x\times 2a+b\right)^{2}}{\left(2a\right)^{2}} as a single fraction.

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

Use binomial theorem \left(p+q\right)^{2}=p^{2}+2pq+q^{2} to expand \left(x\times 2a+b\right)^{2}.

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

Expand \left(2a\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

Cancel out 4 in both numerator and denominator.

$$\frac{4a^{2}x^{2}+4abx+b^{2}}{a^{2}}a^{2}=b^{2}-4ac$$

Express \frac{4a^{2}x^{2}+4abx+b^{2}}{a^{2}}a^{2} as a single fraction.

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

Cancel out a^{2} in both numerator and denominator.

$$4a^{2}x^{2}+4abx+b^{2}=b^{2}-4ac$$

Subtract b^{2} from both sides.

$$4a^{2}x^{2}+4abx+b^{2}-b^{2}=-4ac$$

Combine b^{2} and -b^{2} to get 0.

$$4a^{2}x^{2}+4abx=-4ac$$

Subtract 4a^{2}x^{2} from both sides.

$$4abx=-4ac-4a^{2}x^{2}$$

The equation is in standard form.

$$4axb=-4a^{2}x^{2}-4ac$$

Divide both sides by 4ax.

$$\frac{4axb}{4ax}=-\frac{4a\left(ax^{2}+c\right)}{4ax}$$

Dividing by 4ax undoes the multiplication by 4ax.

$$b=-\frac{4a\left(ax^{2}+c\right)}{4ax}$$

Divide -4a\left(c+ax^{2}\right) by 4ax.

$$b=-ax-\frac{c}{x}$$

Solve for a

$$\left\{\begin{matrix}a=-\frac{bx+c}{x^{2}}\text{, }&\left(c\neq 0\text{ or }b\neq 0\right)\text{ and }\left(b=0\text{ or }x\neq -\frac{c}{b}\right)\text{ and }x\neq 0\text{ and }c\neq -bx\\a\neq 0\text{, }&c=0\text{ and }x=0\end{matrix}\right.$$