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

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

Solve for a

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

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.

$$-b^{2}+4ac=-8a$$

Add 8a to both sides.

$$-b^{2}+4ac+8a=0$$

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

$$4ac+8a=b^{2}$$

Combine all terms containing a.

$$\left(4c+8\right)a=b^{2}$$

Divide both sides by 4c+8.

$$\frac{\left(4c+8\right)a}{4c+8}=\frac{b^{2}}{4c+8}$$

Dividing by 4c+8 undoes the multiplication by 4c+8.

$$a=\frac{b^{2}}{4c+8}$$

Divide b^{2} by 4c+8.

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

Variable a cannot be equal to 0.

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

Solve for b

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

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