Solve b^2-4ac=3

$$b ^ { 2 } - 4 a c = 3$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{b^{2}-3}{4c}\text{, }&c\neq 0\\a\in \mathrm{C}\text{, }&\left(b=\sqrt{3}\text{ or }b=-\sqrt{3}\right)\text{ and }c=0\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract b^{2} from both sides.

$$-4ac=3-b^{2}$$

The equation is in standard form.

$$\left(-4c\right)a=3-b^{2}$$

Divide both sides by -4c.

$$\frac{\left(-4c\right)a}{-4c}=\frac{3-b^{2}}{-4c}$$

Dividing by -4c undoes the multiplication by -4c.

$$a=\frac{3-b^{2}}{-4c}$$

Divide -b^{2}+3 by -4c.

$$a=-\frac{3-b^{2}}{4c}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{b^{2}-3}{4c}\text{, }&c\neq 0\\a\in \mathrm{R}\text{, }&c=0\text{ and }|b|=\sqrt{3}\end{matrix}\right.$$

Steps for Solving Linear Equation

Subtract b^{2} from both sides.

$$-4ac=3-b^{2}$$

The equation is in standard form.

$$\left(-4c\right)a=3-b^{2}$$

Divide both sides by -4c.

$$\frac{\left(-4c\right)a}{-4c}=\frac{3-b^{2}}{-4c}$$

Dividing by -4c undoes the multiplication by -4c.

$$a=\frac{3-b^{2}}{-4c}$$

Divide 3-b^{2} by -4c.

$$a=-\frac{3-b^{2}}{4c}$$

Solve for b (complex solution)

$$b=-\sqrt{4ac+3}$$

$$b=\sqrt{4ac+3}$$

Solve for b

$$b=\sqrt{4ac+3}$$

$$b=-\sqrt{4ac+3}\text{, }\left(c\leq 0\text{ or }a\geq -\frac{3}{4c}\right)\text{ and }\left(c\geq 0\text{ or }a\leq -\frac{3}{4c}\right)$$