Solve 300/x-300/(x-5)=5

$$\frac{ 300 }{ x } - \frac{ 300 }{ (x-5) } =5$$

Solve for x (complex solution)

$$x=\frac{-5\sqrt{47}i+5}{2}\approx 2.5-17.139136501i$$

$$x=\frac{5+5\sqrt{47}i}{2}\approx 2.5+17.139136501i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,5 since division by zero is not defined. Multiply both sides of the equation by x\left(x-5\right), the least common multiple of x,x-5.

$$\left(x-5\right)\times 300-x\times 300=5x\left(x-5\right)$$

Use the distributive property to multiply x-5 by 300.

$$300x-1500-x\times 300=5x\left(x-5\right)$$

Use the distributive property to multiply 5x by x-5.

$$300x-1500-x\times 300=5x^{2}-25x$$

Subtract 5x^{2} from both sides.

$$300x-1500-x\times 300-5x^{2}=-25x$$

Add 25x to both sides.

$$300x-1500-x\times 300-5x^{2}+25x=0$$

Combine 300x and 25x to get 325x.

$$325x-1500-x\times 300-5x^{2}=0$$

Multiply -1 and 300 to get -300.

$$325x-1500-300x-5x^{2}=0$$

Combine 325x and -300x to get 25x.

$$25x-1500-5x^{2}=0$$

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.

$$-5x^{2}+25x-1500=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 25 for b, and -1500 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-5\right)\left(-1500\right)}}{2\left(-5\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-5\right)\left(-1500\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-25±\sqrt{625+20\left(-1500\right)}}{2\left(-5\right)}$$

Multiply 20 times -1500.

$$x=\frac{-25±\sqrt{625-30000}}{2\left(-5\right)}$$

Add 625 to -30000.

$$x=\frac{-25±\sqrt{-29375}}{2\left(-5\right)}$$

Take the square root of -29375.

$$x=\frac{-25±25\sqrt{47}i}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-25±25\sqrt{47}i}{-10}$$

Now solve the equation x=\frac{-25±25\sqrt{47}i}{-10} when ± is plus. Add -25 to 25i\sqrt{47}.

$$x=\frac{-25+25\sqrt{47}i}{-10}$$

Divide -25+25i\sqrt{47} by -10.

$$x=\frac{-5\sqrt{47}i+5}{2}$$

Now solve the equation x=\frac{-25±25\sqrt{47}i}{-10} when ± is minus. Subtract 25i\sqrt{47} from -25.

$$x=\frac{-25\sqrt{47}i-25}{-10}$$

Divide -25-25i\sqrt{47} by -10.

$$x=\frac{5+5\sqrt{47}i}{2}$$

The equation is now solved.

$$x=\frac{-5\sqrt{47}i+5}{2} x=\frac{5+5\sqrt{47}i}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,5 since division by zero is not defined. Multiply both sides of the equation by x\left(x-5\right), the least common multiple of x,x-5.

$$\left(x-5\right)\times 300-x\times 300=5x\left(x-5\right)$$

Use the distributive property to multiply x-5 by 300.

$$300x-1500-x\times 300=5x\left(x-5\right)$$

Use the distributive property to multiply 5x by x-5.

$$300x-1500-x\times 300=5x^{2}-25x$$

Subtract 5x^{2} from both sides.

$$300x-1500-x\times 300-5x^{2}=-25x$$

Add 25x to both sides.

$$300x-1500-x\times 300-5x^{2}+25x=0$$

Combine 300x and 25x to get 325x.

$$325x-1500-x\times 300-5x^{2}=0$$

Add 1500 to both sides. Anything plus zero gives itself.

$$325x-x\times 300-5x^{2}=1500$$

Multiply -1 and 300 to get -300.

$$325x-300x-5x^{2}=1500$$

Combine 325x and -300x to get 25x.

$$25x-5x^{2}=1500$$

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.

$$-5x^{2}+25x=1500$$

Divide both sides by -5.

$$\frac{-5x^{2}+25x}{-5}=\frac{1500}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{25}{-5}x=\frac{1500}{-5}$$

Divide 25 by -5.

$$x^{2}-5x=\frac{1500}{-5}$$

Divide 1500 by -5.

$$x^{2}-5x=-300$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-300+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=-300+\frac{25}{4}$$

Add -300 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=-\frac{1175}{4}$$

Factor x^{2}-5x+\frac{25}{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}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{1175}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{1175}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{5\sqrt{47}i}{2} x-\frac{5}{2}=-\frac{5\sqrt{47}i}{2}$$

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

$$x=\frac{5+5\sqrt{47}i}{2} x=\frac{-5\sqrt{47}i+5}{2}$$