Solve y=-ax^2+bx+c

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

Solve for a

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

Steps for Solving Linear Equation

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

$$\left(-a\right)x^{2}+bx+c=y$$

Subtract bx from both sides.

$$\left(-a\right)x^{2}+c=y-bx$$

Subtract c from both sides.

$$\left(-a\right)x^{2}=y-bx-c$$

Reorder the terms.

$$-ax^{2}=-bx+y-c$$

The equation is in standard form.

$$\left(-x^{2}\right)a=-bx+y-c$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)a}{-x^{2}}=\frac{-bx+y-c}{-x^{2}}$$

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

$$a=\frac{-bx+y-c}{-x^{2}}$$

Divide y-bx-c by -x^{2}.

$$a=-\frac{-bx+y-c}{x^{2}}$$

Solve for b

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

Steps for Solving Linear Equation

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

$$\left(-a\right)x^{2}+bx+c=y$$

Subtract \left(-a\right)x^{2} from both sides.

$$bx+c=y-\left(-a\right)x^{2}$$

Subtract c from both sides.

$$bx=y-\left(-a\right)x^{2}-c$$

Multiply -1 and -1 to get 1.

$$bx=y+ax^{2}-c$$

The equation is in standard form.

$$xb=ax^{2}+y-c$$

Divide both sides by x.

$$\frac{xb}{x}=\frac{ax^{2}+y-c}{x}$$

Dividing by x undoes the multiplication by x.

$$b=\frac{ax^{2}+y-c}{x}$$