Solve (50*(1-1/3x)-30)*10*(1+5/2x)=250

$$(50 \times (1- \frac{ 1 }{ 3 } x)-30) \times 10 \times (1+ \frac{ 5 }{ 2 } x)=250$$

Solve for x

$$x=\frac{1}{5}=0.2$$

$$x=\frac{3}{5}=0.6$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 50 by 1-\frac{1}{3}x.

$$\left(50+50\left(-\frac{1}{3}\right)x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Express 50\left(-\frac{1}{3}\right) as a single fraction.

$$\left(50+\frac{50\left(-1\right)}{3}x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Multiply 50 and -1 to get -50.

$$\left(50+\frac{-50}{3}x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Fraction \frac{-50}{3} can be rewritten as -\frac{50}{3} by extracting the negative sign.

$$\left(50-\frac{50}{3}x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Subtract 30 from 50 to get 20.

$$\left(20-\frac{50}{3}x\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Use the distributive property to multiply 20-\frac{50}{3}x by 10.

$$\left(200-\frac{50}{3}x\times 10\right)\left(1+\frac{5}{2}x\right)=250$$

Express -\frac{50}{3}\times 10 as a single fraction.

$$\left(200+\frac{-50\times 10}{3}x\right)\left(1+\frac{5}{2}x\right)=250$$

Multiply -50 and 10 to get -500.

$$\left(200+\frac{-500}{3}x\right)\left(1+\frac{5}{2}x\right)=250$$

Fraction \frac{-500}{3} can be rewritten as -\frac{500}{3} by extracting the negative sign.

$$\left(200-\frac{500}{3}x\right)\left(1+\frac{5}{2}x\right)=250$$

Apply the distributive property by multiplying each term of 200-\frac{500}{3}x by each term of 1+\frac{5}{2}x.

$$200+200\times \frac{5}{2}x-\frac{500}{3}x-\frac{500}{3}x\times \frac{5}{2}x=250$$

Multiply x and x to get x^{2}.

$$200+200\times \frac{5}{2}x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Express 200\times \frac{5}{2} as a single fraction.

$$200+\frac{200\times 5}{2}x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Multiply 200 and 5 to get 1000.

$$200+\frac{1000}{2}x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Divide 1000 by 2 to get 500.

$$200+500x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Combine 500x and -\frac{500}{3}x to get \frac{1000}{3}x.

$$200+\frac{1000}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Multiply -\frac{500}{3} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator.

$$200+\frac{1000}{3}x+\frac{-500\times 5}{3\times 2}x^{2}=250$$

Do the multiplications in the fraction \frac{-500\times 5}{3\times 2}.

$$200+\frac{1000}{3}x+\frac{-2500}{6}x^{2}=250$$

Reduce the fraction \frac{-2500}{6} to lowest terms by extracting and canceling out 2.

$$200+\frac{1000}{3}x-\frac{1250}{3}x^{2}=250$$

Subtract 250 from both sides.

$$200+\frac{1000}{3}x-\frac{1250}{3}x^{2}-250=0$$

Subtract 250 from 200 to get -50.

$$-50+\frac{1000}{3}x-\frac{1250}{3}x^{2}=0$$

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.

$$-\frac{1250}{3}x^{2}+\frac{1000}{3}x-50=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1250}{3} for a, \frac{1000}{3} for b, and -50 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\frac{1000}{3}±\sqrt{\left(\frac{1000}{3}\right)^{2}-4\left(-\frac{1250}{3}\right)\left(-50\right)}}{2\left(-\frac{1250}{3}\right)}$$

Square \frac{1000}{3} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\frac{1000}{3}±\sqrt{\frac{1000000}{9}-4\left(-\frac{1250}{3}\right)\left(-50\right)}}{2\left(-\frac{1250}{3}\right)}$$

Multiply -4 times -\frac{1250}{3}.

$$x=\frac{-\frac{1000}{3}±\sqrt{\frac{1000000}{9}+\frac{5000}{3}\left(-50\right)}}{2\left(-\frac{1250}{3}\right)}$$

Multiply \frac{5000}{3} times -50.

$$x=\frac{-\frac{1000}{3}±\sqrt{\frac{1000000}{9}-\frac{250000}{3}}}{2\left(-\frac{1250}{3}\right)}$$

Add \frac{1000000}{9} to -\frac{250000}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=\frac{-\frac{1000}{3}±\sqrt{\frac{250000}{9}}}{2\left(-\frac{1250}{3}\right)}$$

Take the square root of \frac{250000}{9}.

$$x=\frac{-\frac{1000}{3}±\frac{500}{3}}{2\left(-\frac{1250}{3}\right)}$$

Multiply 2 times -\frac{1250}{3}.

$$x=\frac{-\frac{1000}{3}±\frac{500}{3}}{-\frac{2500}{3}}$$

Now solve the equation x=\frac{-\frac{1000}{3}±\frac{500}{3}}{-\frac{2500}{3}} when ± is plus. Add -\frac{1000}{3} to \frac{500}{3} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{\frac{500}{3}}{-\frac{2500}{3}}$$

Divide -\frac{500}{3} by -\frac{2500}{3} by multiplying -\frac{500}{3} by the reciprocal of -\frac{2500}{3}.

$$x=\frac{1}{5}$$

Now solve the equation x=\frac{-\frac{1000}{3}±\frac{500}{3}}{-\frac{2500}{3}} when ± is minus. Subtract \frac{500}{3} from -\frac{1000}{3} by finding a common denominator and subtracting the numerators. Then reduce the fraction to lowest terms if possible.

$$x=-\frac{500}{-\frac{2500}{3}}$$

Divide -500 by -\frac{2500}{3} by multiplying -500 by the reciprocal of -\frac{2500}{3}.

$$x=\frac{3}{5}$$

The equation is now solved.

$$x=\frac{1}{5} x=\frac{3}{5}$$

Steps for Completing the Square

Use the distributive property to multiply 50 by 1-\frac{1}{3}x.

$$\left(50+50\left(-\frac{1}{3}\right)x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Express 50\left(-\frac{1}{3}\right) as a single fraction.

$$\left(50+\frac{50\left(-1\right)}{3}x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Multiply 50 and -1 to get -50.

$$\left(50+\frac{-50}{3}x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Fraction \frac{-50}{3} can be rewritten as -\frac{50}{3} by extracting the negative sign.

$$\left(50-\frac{50}{3}x-30\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Subtract 30 from 50 to get 20.

$$\left(20-\frac{50}{3}x\right)\times 10\left(1+\frac{5}{2}x\right)=250$$

Use the distributive property to multiply 20-\frac{50}{3}x by 10.

$$\left(200-\frac{50}{3}x\times 10\right)\left(1+\frac{5}{2}x\right)=250$$

Express -\frac{50}{3}\times 10 as a single fraction.

$$\left(200+\frac{-50\times 10}{3}x\right)\left(1+\frac{5}{2}x\right)=250$$

Multiply -50 and 10 to get -500.

$$\left(200+\frac{-500}{3}x\right)\left(1+\frac{5}{2}x\right)=250$$

Fraction \frac{-500}{3} can be rewritten as -\frac{500}{3} by extracting the negative sign.

$$\left(200-\frac{500}{3}x\right)\left(1+\frac{5}{2}x\right)=250$$

Apply the distributive property by multiplying each term of 200-\frac{500}{3}x by each term of 1+\frac{5}{2}x.

$$200+200\times \frac{5}{2}x-\frac{500}{3}x-\frac{500}{3}x\times \frac{5}{2}x=250$$

Multiply x and x to get x^{2}.

$$200+200\times \frac{5}{2}x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Express 200\times \frac{5}{2} as a single fraction.

$$200+\frac{200\times 5}{2}x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Multiply 200 and 5 to get 1000.

$$200+\frac{1000}{2}x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Divide 1000 by 2 to get 500.

$$200+500x-\frac{500}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Combine 500x and -\frac{500}{3}x to get \frac{1000}{3}x.

$$200+\frac{1000}{3}x-\frac{500}{3}x^{2}\times \frac{5}{2}=250$$

Multiply -\frac{500}{3} times \frac{5}{2} by multiplying numerator times numerator and denominator times denominator.

$$200+\frac{1000}{3}x+\frac{-500\times 5}{3\times 2}x^{2}=250$$

Do the multiplications in the fraction \frac{-500\times 5}{3\times 2}.

$$200+\frac{1000}{3}x+\frac{-2500}{6}x^{2}=250$$

Reduce the fraction \frac{-2500}{6} to lowest terms by extracting and canceling out 2.

$$200+\frac{1000}{3}x-\frac{1250}{3}x^{2}=250$$

Subtract 200 from both sides.

$$\frac{1000}{3}x-\frac{1250}{3}x^{2}=250-200$$

Subtract 200 from 250 to get 50.

$$\frac{1000}{3}x-\frac{1250}{3}x^{2}=50$$

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.

$$-\frac{1250}{3}x^{2}+\frac{1000}{3}x=50$$

Divide both sides of the equation by -\frac{1250}{3}, which is the same as multiplying both sides by the reciprocal of the fraction.

$$\frac{-\frac{1250}{3}x^{2}+\frac{1000}{3}x}{-\frac{1250}{3}}=\frac{50}{-\frac{1250}{3}}$$

Dividing by -\frac{1250}{3} undoes the multiplication by -\frac{1250}{3}.

$$x^{2}+\frac{\frac{1000}{3}}{-\frac{1250}{3}}x=\frac{50}{-\frac{1250}{3}}$$

Divide \frac{1000}{3} by -\frac{1250}{3} by multiplying \frac{1000}{3} by the reciprocal of -\frac{1250}{3}.

$$x^{2}-\frac{4}{5}x=\frac{50}{-\frac{1250}{3}}$$

Divide 50 by -\frac{1250}{3} by multiplying 50 by the reciprocal of -\frac{1250}{3}.

$$x^{2}-\frac{4}{5}x=-\frac{3}{25}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=-\frac{3}{25}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{-3+4}{25}$$

Add -\frac{3}{25} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{1}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{1}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{1}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{1}{5} x-\frac{2}{5}=-\frac{1}{5}$$

Add \frac{2}{5} to both sides of the equation.

$$x=\frac{3}{5} x=\frac{1}{5}$$