Solve 0=ax^2+5ax+4

$$0 = a x ^ { 2 } + 5 a x + 4$$

Solve for a

$$a=-\frac{4}{x\left(x+5\right)}$$

$$x\neq -5\text{ and }x\neq 0$$

Steps for Solving Linear Equation

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

$$ax^{2}+5ax+4=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$ax^{2}+5ax=-4$$

Combine all terms containing a.

$$\left(x^{2}+5x\right)a=-4$$

Divide both sides by x^{2}+5x.

$$\frac{\left(x^{2}+5x\right)a}{x^{2}+5x}=-\frac{4}{x^{2}+5x}$$

Dividing by x^{2}+5x undoes the multiplication by x^{2}+5x.

$$a=-\frac{4}{x^{2}+5x}$$

Divide -4 by x^{2}+5x.

$$a=-\frac{4}{x\left(x+5\right)}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{25a^{2}-16a}}{2a}-\frac{5}{2}$$

$$x=-\frac{\sqrt{25a^{2}-16a}}{2a}-\frac{5}{2}\text{, }a\neq 0$$

Solve for x

$$x=\frac{\sqrt{25a^{2}-16a}}{2a}-\frac{5}{2}$$

$$x=-\frac{\sqrt{25a^{2}-16a}}{2a}-\frac{5}{2}\text{, }a<0\text{ or }a\geq \frac{16}{25}$$