Solve 16x^2-yx+100=0

$$16 { x }^{ 2 } -yx+100=0$$

Solve for y

$$y=16x+\frac{100}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 16x^{2} from both sides. Anything subtracted from zero gives its negation.

$$-yx+100=-16x^{2}$$

Subtract 100 from both sides.

$$-yx=-16x^{2}-100$$

The equation is in standard form.

$$\left(-x\right)y=-16x^{2}-100$$

Divide both sides by -x.

$$\frac{\left(-x\right)y}{-x}=\frac{-16x^{2}-100}{-x}$$

Dividing by -x undoes the multiplication by -x.

$$y=\frac{-16x^{2}-100}{-x}$$

Divide -16x^{2}-100 by -x.

$$y=16x+\frac{100}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{y^{2}-6400}+y}{32}$$

$$x=\frac{-\sqrt{y^{2}-6400}+y}{32}$$

Solve for x

$$x=\frac{\sqrt{y^{2}-6400}+y}{32}$$

$$x=\frac{-\sqrt{y^{2}-6400}+y}{32}\text{, }|y|\geq 80$$