Solve 0.5x^2+0.5x=200

$$0.5 { x }^{ 2 } +0.5x = 200$$

Solve for x

$$x = \frac{\sqrt{1601} - 1}{2} \approx 19.506249024$$

$$x=\frac{-\sqrt{1601}-1}{2}\approx -20.506249024$$

Steps Using the Quadratic Formula

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.

$$\frac{1}{2}x^{2}+\frac{1}{2}x=200$$

Subtract 200 from both sides of the equation.

$$\frac{1}{2}x^{2}+\frac{1}{2}x-200=200-200$$

Subtracting 200 from itself leaves 0.

$$\frac{1}{2}x^{2}+\frac{1}{2}x-200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{1}{2} for a, \frac{1}{2} for b, and -200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\times \frac{1}{2}\left(-200\right)}}{2\times \frac{1}{2}}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\times \frac{1}{2}\left(-200\right)}}{2\times \frac{1}{2}}$$

Multiply -4 times \frac{1}{2}.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-2\left(-200\right)}}{2\times \frac{1}{2}}$$

Multiply -2 times -200.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+400}}{2\times \frac{1}{2}}$$

Add \frac{1}{4} to 400.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1601}{4}}}{2\times \frac{1}{2}}$$

Take the square root of \frac{1601}{4}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{1601}}{2}}{2\times \frac{1}{2}}$$

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

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{1601}}{2}}{1}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{1601}}{2}}{1} when ± is plus. Add -\frac{1}{2} to \frac{\sqrt{1601}}{2}.

$$x=\frac{\sqrt{1601}-1}{2}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{1601}}{2}}{1} when ± is minus. Subtract \frac{\sqrt{1601}}{2} from -\frac{1}{2}.

$$x=\frac{-\sqrt{1601}-1}{2}$$

The equation is now solved.

$$x=\frac{\sqrt{1601}-1}{2} x=\frac{-\sqrt{1601}-1}{2}$$

Steps for Completing the Square

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.

$$\frac{1}{2}x^{2}+\frac{1}{2}x=200$$

Multiply both sides by 2.

$$\frac{\frac{1}{2}x^{2}+\frac{1}{2}x}{\frac{1}{2}}=\frac{200}{\frac{1}{2}}$$

Dividing by \frac{1}{2} undoes the multiplication by \frac{1}{2}.

$$x^{2}+\frac{\frac{1}{2}}{\frac{1}{2}}x=\frac{200}{\frac{1}{2}}$$

Divide \frac{1}{2} by \frac{1}{2} by multiplying \frac{1}{2} by the reciprocal of \frac{1}{2}.

$$x^{2}+x=\frac{200}{\frac{1}{2}}$$

Divide 200 by \frac{1}{2} by multiplying 200 by the reciprocal of \frac{1}{2}.

$$x^{2}+x=400$$

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=400+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=400+\frac{1}{4}$$

Add 400 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{1601}{4}$$

Factor x^{2}+x+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{1601}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{1601}{4}}$$

Simplify.

$$x+\frac{1}{2}=\frac{\sqrt{1601}}{2} x+\frac{1}{2}=-\frac{\sqrt{1601}}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$x=\frac{\sqrt{1601}-1}{2} x=\frac{-\sqrt{1601}-1}{2}$$