Solve 420/x-5-5=420/x

$$\frac { 420 } { x - 5 } - 5 = \frac { 420 } { x }$$

Solve for x

$$x = \frac{\sqrt{1705} + 5}{2} \approx 23.145822822$$

$$x=\frac{5-\sqrt{1705}}{2}\approx -18.145822822$$

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-5,x.

$$x\times 420+x\left(x-5\right)\left(-5\right)=\left(x-5\right)\times 420$$

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

$$x\times 420+\left(x^{2}-5x\right)\left(-5\right)=\left(x-5\right)\times 420$$

Use the distributive property to multiply x^{2}-5x by -5.

$$x\times 420-5x^{2}+25x=\left(x-5\right)\times 420$$

Combine x\times 420 and 25x to get 445x.

$$445x-5x^{2}=\left(x-5\right)\times 420$$

Use the distributive property to multiply x-5 by 420.

$$445x-5x^{2}=420x-2100$$

Subtract 420x from both sides.

$$445x-5x^{2}-420x=-2100$$

Combine 445x and -420x to get 25x.

$$25x-5x^{2}=-2100$$

Add 2100 to both sides.

$$25x-5x^{2}+2100=0$$

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.

$$-5x^{2}+25x+2100=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 25 for b, and 2100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-25±\sqrt{25^{2}-4\left(-5\right)\times 2100}}{2\left(-5\right)}$$

Square 25.

$$x=\frac{-25±\sqrt{625-4\left(-5\right)\times 2100}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-25±\sqrt{625+20\times 2100}}{2\left(-5\right)}$$

Multiply 20 times 2100.

$$x=\frac{-25±\sqrt{625+42000}}{2\left(-5\right)}$$

Add 625 to 42000.

$$x=\frac{-25±\sqrt{42625}}{2\left(-5\right)}$$

Take the square root of 42625.

$$x=\frac{-25±5\sqrt{1705}}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-25±5\sqrt{1705}}{-10}$$

Now solve the equation x=\frac{-25±5\sqrt{1705}}{-10} when ± is plus. Add -25 to 5\sqrt{1705}.

$$x=\frac{5\sqrt{1705}-25}{-10}$$

Divide -25+5\sqrt{1705} by -10.

$$x=\frac{5-\sqrt{1705}}{2}$$

Now solve the equation x=\frac{-25±5\sqrt{1705}}{-10} when ± is minus. Subtract 5\sqrt{1705} from -25.

$$x=\frac{-5\sqrt{1705}-25}{-10}$$

Divide -25-5\sqrt{1705} by -10.

$$x=\frac{\sqrt{1705}+5}{2}$$

The equation is now solved.

$$x=\frac{5-\sqrt{1705}}{2} x=\frac{\sqrt{1705}+5}{2}$$

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-5,x.

$$x\times 420+x\left(x-5\right)\left(-5\right)=\left(x-5\right)\times 420$$

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

$$x\times 420+\left(x^{2}-5x\right)\left(-5\right)=\left(x-5\right)\times 420$$

Use the distributive property to multiply x^{2}-5x by -5.

$$x\times 420-5x^{2}+25x=\left(x-5\right)\times 420$$

Combine x\times 420 and 25x to get 445x.

$$445x-5x^{2}=\left(x-5\right)\times 420$$

Use the distributive property to multiply x-5 by 420.

$$445x-5x^{2}=420x-2100$$

Subtract 420x from both sides.

$$445x-5x^{2}-420x=-2100$$

Combine 445x and -420x to get 25x.

$$25x-5x^{2}=-2100$$

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.

$$-5x^{2}+25x=-2100$$

Divide both sides by -5.

$$\frac{-5x^{2}+25x}{-5}=-\frac{2100}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{25}{-5}x=-\frac{2100}{-5}$$

Divide 25 by -5.

$$x^{2}-5x=-\frac{2100}{-5}$$

Divide -2100 by -5.

$$x^{2}-5x=420$$

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.

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=420+\left(-\frac{5}{2}\right)^{2}$$

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

$$x^{2}-5x+\frac{25}{4}=420+\frac{25}{4}$$

Add 420 to \frac{25}{4}.

$$x^{2}-5x+\frac{25}{4}=\frac{1705}{4}$$

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}.

$$\left(x-\frac{5}{2}\right)^{2}=\frac{1705}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{\frac{1705}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{1705}}{2} x-\frac{5}{2}=-\frac{\sqrt{1705}}{2}$$

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

$$x=\frac{\sqrt{1705}+5}{2} x=\frac{5-\sqrt{1705}}{2}$$