Solve -b^2+4ac/4a=-3

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

Solve for a

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

Steps for Solving Linear Equation

Variable a cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 4a.

$$-b^{2}+4ac=-12a$$

Add 12a to both sides.

$$-b^{2}+4ac+12a=0$$

Add b^{2} to both sides. Anything plus zero gives itself.

$$4ac+12a=b^{2}$$

Combine all terms containing a.

$$\left(4c+12\right)a=b^{2}$$

Divide both sides by 4c+12.

$$\frac{\left(4c+12\right)a}{4c+12}=\frac{b^{2}}{4c+12}$$

Dividing by 4c+12 undoes the multiplication by 4c+12.

$$a=\frac{b^{2}}{4c+12}$$

Divide b^{2} by 4c+12.

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

Variable a cannot be equal to 0.

$$a=\frac{b^{2}}{4\left(c+3\right)}\text{, }a\neq 0$$

Solve for b

$$b=2\sqrt{a\left(c+3\right)}$$

$$b=-2\sqrt{a\left(c+3\right)}\text{, }\left(c\geq -3\text{ and }a>0\right)\text{ or }\left(c\leq -3\text{ and }a<0\right)$$