Solve 115=xleft(1+3x10^-3right).50

$$115 = x \left( 1+3x { 10 }^{ -3 } \right) .50$$

Solve for x

$$x = \frac{100 \sqrt{94} - 500}{3} \approx 156.511990494$$

$$x=\frac{-100\sqrt{94}-500}{3}\approx -489.845323828$$

Steps Using the Quadratic Formula

Calculate 10 to the power of -3 and get \frac{1}{1000}.

$$115=x\left(1+3x\times \frac{1}{1000}\right)\times 0.5$$

Multiply 3 and \frac{1}{1000} to get \frac{3}{1000}.

$$115=x\left(1+\frac{3}{1000}x\right)\times 0.5$$

Use the distributive property to multiply x by 1+\frac{3}{1000}x.

$$115=\left(x+\frac{3}{1000}x^{2}\right)\times 0.5$$

Use the distributive property to multiply x+\frac{3}{1000}x^{2} by 0.5.

$$115=0.5x+\frac{3}{2000}x^{2}$$

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

$$0.5x+\frac{3}{2000}x^{2}=115$$

Subtract 115 from both sides.

$$0.5x+\frac{3}{2000}x^{2}-115=0$$

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.

$$\frac{3}{2000}x^{2}+\frac{1}{2}x-115=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute \frac{3}{2000} for a, \frac{1}{2} for b, and -115 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1}{2}±\sqrt{\left(\frac{1}{2}\right)^{2}-4\times \frac{3}{2000}\left(-115\right)}}{2\times \frac{3}{2000}}$$

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

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-4\times \frac{3}{2000}\left(-115\right)}}{2\times \frac{3}{2000}}$$

Multiply -4 times \frac{3}{2000}.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}-\frac{3}{500}\left(-115\right)}}{2\times \frac{3}{2000}}$$

Multiply -\frac{3}{500} times -115.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{1}{4}+\frac{69}{100}}}{2\times \frac{3}{2000}}$$

Add \frac{1}{4} to \frac{69}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1}{2}±\sqrt{\frac{47}{50}}}{2\times \frac{3}{2000}}$$

Take the square root of \frac{47}{50}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{94}}{10}}{2\times \frac{3}{2000}}$$

Multiply 2 times \frac{3}{2000}.

$$x=\frac{-\frac{1}{2}±\frac{\sqrt{94}}{10}}{\frac{3}{1000}}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{94}}{10}}{\frac{3}{1000}} when ± is plus. Add -\frac{1}{2} to \frac{\sqrt{94}}{10}.

$$x=\frac{\frac{\sqrt{94}}{10}-\frac{1}{2}}{\frac{3}{1000}}$$

Divide -\frac{1}{2}+\frac{\sqrt{94}}{10} by \frac{3}{1000} by multiplying -\frac{1}{2}+\frac{\sqrt{94}}{10} by the reciprocal of \frac{3}{1000}.

$$x=\frac{100\sqrt{94}-500}{3}$$

Now solve the equation x=\frac{-\frac{1}{2}±\frac{\sqrt{94}}{10}}{\frac{3}{1000}} when ± is minus. Subtract \frac{\sqrt{94}}{10} from -\frac{1}{2}.

$$x=\frac{-\frac{\sqrt{94}}{10}-\frac{1}{2}}{\frac{3}{1000}}$$

Divide -\frac{1}{2}-\frac{\sqrt{94}}{10} by \frac{3}{1000} by multiplying -\frac{1}{2}-\frac{\sqrt{94}}{10} by the reciprocal of \frac{3}{1000}.

$$x=\frac{-100\sqrt{94}-500}{3}$$

The equation is now solved.

$$x=\frac{100\sqrt{94}-500}{3} x=\frac{-100\sqrt{94}-500}{3}$$

Steps for Completing the Square

Calculate 10 to the power of -3 and get \frac{1}{1000}.

$$115=x\left(1+3x\times \frac{1}{1000}\right)\times 0.5$$

Multiply 3 and \frac{1}{1000} to get \frac{3}{1000}.

$$115=x\left(1+\frac{3}{1000}x\right)\times 0.5$$

Use the distributive property to multiply x by 1+\frac{3}{1000}x.

$$115=\left(x+\frac{3}{1000}x^{2}\right)\times 0.5$$

Use the distributive property to multiply x+\frac{3}{1000}x^{2} by 0.5.

$$115=0.5x+\frac{3}{2000}x^{2}$$

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

$$0.5x+\frac{3}{2000}x^{2}=115$$

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.

$$\frac{3}{2000}x^{2}+\frac{1}{2}x=115$$

Divide both sides of the equation by \frac{3}{2000}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{\frac{3}{2000}x^{2}+\frac{1}{2}x}{\frac{3}{2000}}=\frac{115}{\frac{3}{2000}}$$

Dividing by \frac{3}{2000} undoes the multiplication by \frac{3}{2000}.

$$x^{2}+\frac{\frac{1}{2}}{\frac{3}{2000}}x=\frac{115}{\frac{3}{2000}}$$

Divide \frac{1}{2} by \frac{3}{2000} by multiplying \frac{1}{2} by the reciprocal of \frac{3}{2000}.

$$x^{2}+\frac{1000}{3}x=\frac{115}{\frac{3}{2000}}$$

Divide 115 by \frac{3}{2000} by multiplying 115 by the reciprocal of \frac{3}{2000}.

$$x^{2}+\frac{1000}{3}x=\frac{230000}{3}$$

Divide \frac{1000}{3}, the coefficient of the x term, by 2 to get \frac{500}{3}. Then add the square of \frac{500}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1000}{3}x+\left(\frac{500}{3}\right)^{2}=\frac{230000}{3}+\left(\frac{500}{3}\right)^{2}$$

Square \frac{500}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1000}{3}x+\frac{250000}{9}=\frac{230000}{3}+\frac{250000}{9}$$

Add \frac{230000}{3} to \frac{250000}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1000}{3}x+\frac{250000}{9}=\frac{940000}{9}$$

Factor x^{2}+\frac{1000}{3}x+\frac{250000}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{500}{3}\right)^{2}=\frac{940000}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{500}{3}\right)^{2}}=\sqrt{\frac{940000}{9}}$$

Simplify.

$$x+\frac{500}{3}=\frac{100\sqrt{94}}{3} x+\frac{500}{3}=-\frac{100\sqrt{94}}{3}$$

Subtract \frac{500}{3} from both sides of the equation.

$$x=\frac{100\sqrt{94}-500}{3} x=\frac{-100\sqrt{94}-500}{3}$$