Solve Y=-ax^2+0.1b+c

$$Y = - a x ^ { 2 } + 0.1 b + c$$

Solve for a (complex solution)

$$\left\{\begin{matrix}a=\frac{b+10c-10Y}{10x^{2}}\text{, }&x\neq 0\\a\in \mathrm{C}\text{, }&Y=\frac{b}{10}+c\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$\left(-a\right)x^{2}+0.1b+c=Y$$

Subtract 0.1b from both sides.

$$\left(-a\right)x^{2}+c=Y-0.1b$$

Subtract c from both sides.

$$\left(-a\right)x^{2}=Y-0.1b-c$$

Reorder the terms.

$$-ax^{2}=Y-c-0.1b$$

The equation is in standard form.

$$\left(-x^{2}\right)a=-\frac{b}{10}+Y-c$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)a}{-x^{2}}=\frac{-\frac{b}{10}+Y-c}{-x^{2}}$$

Dividing by -x^{2} undoes the multiplication by -x^{2}.

$$a=\frac{-\frac{b}{10}+Y-c}{-x^{2}}$$

Divide Y-\frac{b}{10}-c by -x^{2}.

$$a=-\frac{10Y-b-10c}{10x^{2}}$$

Solve for a

$$\left\{\begin{matrix}a=\frac{b+10c-10Y}{10x^{2}}\text{, }&x\neq 0\\a\in \mathrm{R}\text{, }&Y=\frac{b}{10}+c\text{ and }x=0\end{matrix}\right.$$

Steps for Solving Linear Equation

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

$$\left(-a\right)x^{2}+0.1b+c=Y$$

Subtract 0.1b from both sides.

$$\left(-a\right)x^{2}+c=Y-0.1b$$

Subtract c from both sides.

$$\left(-a\right)x^{2}=Y-0.1b-c$$

Reorder the terms.

$$-ax^{2}=Y-c-0.1b$$

The equation is in standard form.

$$\left(-x^{2}\right)a=-\frac{b}{10}+Y-c$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)a}{-x^{2}}=\frac{-\frac{b}{10}+Y-c}{-x^{2}}$$

Dividing by -x^{2} undoes the multiplication by -x^{2}.

$$a=\frac{-\frac{b}{10}+Y-c}{-x^{2}}$$

Divide Y-\frac{b}{10}-c by -x^{2}.

$$a=-\frac{10Y-b-10c}{10x^{2}}$$

Solve for Y

$$Y=-ax^{2}+\frac{b}{10}+c$$