Solve x=-1/160x^2+4

$$x = - \frac{ 1 }{ 160 } { x }^{ 2 } +4$$

Solve for x

$$x=8\sqrt{110}-80\approx 3.904707854$$

$$x=-8\sqrt{110}-80\approx -163.904707854$$

Steps Using the Quadratic Formula

Add \frac{1}{160}x^{2} to both sides.

$$x+\frac{1}{160}x^{2}=4$$

Subtract 4 from both sides.

$$x+\frac{1}{160}x^{2}-4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$\frac{1}{160}x^{2}+x-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{160} for a, 1 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times \frac{1}{160}\left(-4\right)}}{2\times \frac{1}{160}}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times \frac{1}{160}\left(-4\right)}}{2\times \frac{1}{160}}$$

Multiply -4 times \frac{1}{160}.

$$x=\frac{-1±\sqrt{1-\frac{1}{40}\left(-4\right)}}{2\times \frac{1}{160}}$$

Multiply -\frac{1}{40} times -4.

$$x=\frac{-1±\sqrt{1+\frac{1}{10}}}{2\times \frac{1}{160}}$$

Add 1 to \frac{1}{10}.

$$x=\frac{-1±\sqrt{\frac{11}{10}}}{2\times \frac{1}{160}}$$

Take the square root of \frac{11}{10}.

$$x=\frac{-1±\frac{\sqrt{110}}{10}}{2\times \frac{1}{160}}$$

Multiply 2 times \frac{1}{160}.

$$x=\frac{-1±\frac{\sqrt{110}}{10}}{\frac{1}{80}}$$

Now solve the equation x=\frac{-1±\frac{\sqrt{110}}{10}}{\frac{1}{80}} when ± is plus. Add -1 to \frac{\sqrt{110}}{10}.

$$x=\frac{\frac{\sqrt{110}}{10}-1}{\frac{1}{80}}$$

Divide -1+\frac{\sqrt{110}}{10} by \frac{1}{80} by multiplying -1+\frac{\sqrt{110}}{10} by the reciprocal of \frac{1}{80}.

$$x=8\sqrt{110}-80$$

Now solve the equation x=\frac{-1±\frac{\sqrt{110}}{10}}{\frac{1}{80}} when ± is minus. Subtract \frac{\sqrt{110}}{10} from -1.

$$x=\frac{-\frac{\sqrt{110}}{10}-1}{\frac{1}{80}}$$

Divide -1-\frac{\sqrt{110}}{10} by \frac{1}{80} by multiplying -1-\frac{\sqrt{110}}{10} by the reciprocal of \frac{1}{80}.

$$x=-8\sqrt{110}-80$$

The equation is now solved.

$$x=8\sqrt{110}-80 x=-8\sqrt{110}-80$$

Steps for Completing the Square

Add \frac{1}{160}x^{2} to both sides.

$$x+\frac{1}{160}x^{2}=4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$\frac{1}{160}x^{2}+x=4$$

Multiply both sides by 160.

$$\frac{\frac{1}{160}x^{2}+x}{\frac{1}{160}}=\frac{4}{\frac{1}{160}}$$

Dividing by \frac{1}{160} undoes the multiplication by \frac{1}{160}.

$$x^{2}+\frac{1}{\frac{1}{160}}x=\frac{4}{\frac{1}{160}}$$

Divide 1 by \frac{1}{160} by multiplying 1 by the reciprocal of \frac{1}{160}.

$$x^{2}+160x=\frac{4}{\frac{1}{160}}$$

Divide 4 by \frac{1}{160} by multiplying 4 by the reciprocal of \frac{1}{160}.

$$x^{2}+160x=640$$

Divide 160, the coefficient of the x term, by 2 to get 80. Then add the square of 80 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+160x+80^{2}=640+80^{2}$$

Square 80.

$$x^{2}+160x+6400=640+6400$$

Add 640 to 6400.

$$x^{2}+160x+6400=7040$$

Factor x^{2}+160x+6400. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+80\right)^{2}=7040$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+80\right)^{2}}=\sqrt{7040}$$

Simplify.

$$x+80=8\sqrt{110} x+80=-8\sqrt{110}$$

Subtract 80 from both sides of the equation.

$$x=8\sqrt{110}-80 x=-8\sqrt{110}-80$$