Solve left(6-aright)x-20=-0.05x^2+10x-40

$$\left( 6-a \right) x-20 = -0.05 { x }^{ 2 } +10x-40$$

Solve for a

$$a=\frac{x}{20}-4+\frac{20}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Use the distributive property to multiply 6-a by x.

$$6x-ax-20=-0.05x^{2}+10x-40$$

Subtract 6x from both sides.

$$-ax-20=-0.05x^{2}+10x-40-6x$$

Combine 10x and -6x to get 4x.

$$-ax-20=-0.05x^{2}+4x-40$$

Add 20 to both sides.

$$-ax=-0.05x^{2}+4x-40+20$$

Add -40 and 20 to get -20.

$$-ax=-0.05x^{2}+4x-20$$

The equation is in standard form.

$$\left(-x\right)a=-\frac{x^{2}}{20}+4x-20$$

Divide both sides by -x.

$$\frac{\left(-x\right)a}{-x}=\frac{-\frac{x^{2}}{20}+4x-20}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$a=\frac{-\frac{x^{2}}{20}+4x-20}{-x}$$

Divide -\frac{x^{2}}{20}+4x-20 by -x.

$$a=\frac{x}{20}-4+\frac{20}{x}$$

Solve for x (complex solution)

$$x=10\sqrt{\left(a+2\right)\left(a+6\right)}+10a+40$$

$$x=-10\sqrt{\left(a+2\right)\left(a+6\right)}+10a+40$$

Solve for x

$$x=10\left(\sqrt{\left(a+2\right)\left(a+6\right)}+a+4\right)$$

$$x=10\left(-\sqrt{\left(a+2\right)\left(a+6\right)}+a+4\right)\text{, }a\leq -6\text{ or }a\geq -2$$