Solve -cx^2+32/2=x(x-5)

$$\frac { - c x ^ { 2 } + 32 } { 2 } = x ( x - 5 )$$

Solve for c

$$c=-2+\frac{10}{x}+\frac{32}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Multiply both sides of the equation by 2.

$$\left(-c\right)x^{2}+32=2x\left(x-5\right)$$

Use the distributive property to multiply 2x by x-5.

$$\left(-c\right)x^{2}+32=2x^{2}-10x$$

Subtract 32 from both sides.

$$\left(-c\right)x^{2}=2x^{2}-10x-32$$

Reorder the terms.

$$-cx^{2}=2x^{2}-10x-32$$

The equation is in standard form.

$$\left(-x^{2}\right)c=2x^{2}-10x-32$$

Divide both sides by -x^{2}.

$$\frac{\left(-x^{2}\right)c}{-x^{2}}=\frac{2x^{2}-10x-32}{-x^{2}}$$

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

$$c=\frac{2x^{2}-10x-32}{-x^{2}}$$

Divide 2x^{2}-10x-32 by -x^{2}.

$$c=-2+\frac{10x+32}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=-\frac{\sqrt{32c+89}-5}{c+2}\text{; }x=\frac{\sqrt{32c+89}+5}{c+2}\text{, }&c\neq -2\\x=-\frac{16}{5}\text{, }&c=-2\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{32c+89}-5}{c+2}\text{; }x=\frac{\sqrt{32c+89}+5}{c+2}\text{, }&c\neq -2\text{ and }c\geq -\frac{89}{32}\\x=-\frac{16}{5}\text{, }&c=-2\end{matrix}\right.$$