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

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

Solve for b

$$\left\{\begin{matrix}b=\frac{-ax^{2}+a^{2}-c}{x}\text{, }&x\neq 0\text{ and }a\neq 0\\b\in \mathrm{R}\text{, }&c=a^{2}\text{ and }x=0\text{ and }a\neq 0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

To multiply powers of the same base, add their exponents. Add 1 and 2 to get 3.

$$a^{3}\times 4=4a^{2}\left(x+\frac{b}{2a}\right)^{2}-\left(b^{2}-4ac\right)$$

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

$$a^{3}\times 4=4a^{2}\left(\frac{x\times 2a}{2a}+\frac{b}{2a}\right)^{2}-\left(b^{2}-4ac\right)$$

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

$$a^{3}\times 4=4a^{2}\times \left(\frac{x\times 2a+b}{2a}\right)^{2}-\left(b^{2}-4ac\right)$$

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

$$a^{3}\times 4=4a^{2}\times \frac{\left(x\times 2a+b\right)^{2}}{\left(2a\right)^{2}}-\left(b^{2}-4ac\right)$$

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

$$a^{3}\times 4=\frac{4\left(x\times 2a+b\right)^{2}}{\left(2a\right)^{2}}a^{2}-\left(b^{2}-4ac\right)$$

To find the opposite of b^{2}-4ac, find the opposite of each term.

$$a^{3}\times 4=\frac{4\left(x\times 2a+b\right)^{2}}{\left(2a\right)^{2}}a^{2}-b^{2}+4ac$$

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

$$a^{3}\times 4=\frac{4\left(4x^{2}a^{2}+4xab+b^{2}\right)}{\left(2a\right)^{2}}a^{2}-b^{2}+4ac$$

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

$$a^{3}\times 4=\frac{4\left(4x^{2}a^{2}+4xab+b^{2}\right)}{2^{2}a^{2}}a^{2}-b^{2}+4ac$$

Calculate 2 to the power of 2 and get 4.

$$a^{3}\times 4=\frac{4\left(4x^{2}a^{2}+4xab+b^{2}\right)}{4a^{2}}a^{2}-b^{2}+4ac$$

Cancel out 4 in both numerator and denominator.

$$a^{3}\times 4=\frac{4a^{2}x^{2}+4abx+b^{2}}{a^{2}}a^{2}-b^{2}+4ac$$

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

$$a^{3}\times 4=\frac{\left(4a^{2}x^{2}+4abx+b^{2}\right)a^{2}}{a^{2}}-b^{2}+4ac$$

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

$$a^{3}\times 4=4a^{2}x^{2}+4abx+b^{2}-b^{2}+4ac$$

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

$$a^{3}\times 4=4a^{2}x^{2}+4abx+4ac$$

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

$$4a^{2}x^{2}+4abx+4ac=a^{3}\times 4$$

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

$$4abx+4ac=a^{3}\times 4-4a^{2}x^{2}$$

Subtract 4ac from both sides.

$$4abx=a^{3}\times 4-4a^{2}x^{2}-4ac$$

The equation is in standard form.

$$4axb=-4a^{2}x^{2}+4a^{3}-4ac$$

Divide both sides by 4ax.

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

Dividing by 4ax undoes the multiplication by 4ax.

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

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

$$b=\frac{-ax^{2}+a^{2}-c}{x}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{-\sqrt{x^{4}+4bx+4c}+x^{2}}{2}\text{, }&\left(c\neq 0\text{ and }b=0\text{ and }c\geq -\frac{x^{4}}{4}\right)\text{ or }\left(b\neq 0\text{ and }x\neq -\frac{c}{b}\text{ and }x^{4}+4bx+4c\geq 0\text{ and }c\geq -\frac{x^{4}}{4}-bx\right)\\a=\frac{\sqrt{x^{4}+4bx+4c}+x^{2}}{2}\text{, }&\left(c\neq 0\text{ or }x\neq 0\right)\text{ and }\left(4c\geq 0\text{ or }x\neq 0\right)\text{ and }x^{4}+4bx+4c\geq 0\text{ and }c\geq -\frac{x^{4}}{4}-bx\end{matrix}\right.$$