Solve 720/x=720/x-10+50

$$\frac { 720 } { x } = \frac { 720 } { x - 10 } + 50$$

Solve for x (complex solution)

$$x=-\sqrt{119}i+5\approx 5-10.908712115i$$

$$x=5+\sqrt{119}i\approx 5+10.908712115i$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x,x-10.

$$\left(x-10\right)\times 720=x\times 720+x\left(x-10\right)\times 50$$

Use the distributive property to multiply x-10 by 720.

$$720x-7200=x\times 720+x\left(x-10\right)\times 50$$

Use the distributive property to multiply x by x-10.

$$720x-7200=x\times 720+\left(x^{2}-10x\right)\times 50$$

Use the distributive property to multiply x^{2}-10x by 50.

$$720x-7200=x\times 720+50x^{2}-500x$$

Combine x\times 720 and -500x to get 220x.

$$720x-7200=220x+50x^{2}$$

Subtract 220x from both sides.

$$720x-7200-220x=50x^{2}$$

Combine 720x and -220x to get 500x.

$$500x-7200=50x^{2}$$

Subtract 50x^{2} from both sides.

$$500x-7200-50x^{2}=0$$

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.

$$-50x^{2}+500x-7200=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -50 for a, 500 for b, and -7200 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-500±\sqrt{500^{2}-4\left(-50\right)\left(-7200\right)}}{2\left(-50\right)}$$

Square 500.

$$x=\frac{-500±\sqrt{250000-4\left(-50\right)\left(-7200\right)}}{2\left(-50\right)}$$

Multiply -4 times -50.

$$x=\frac{-500±\sqrt{250000+200\left(-7200\right)}}{2\left(-50\right)}$$

Multiply 200 times -7200.

$$x=\frac{-500±\sqrt{250000-1440000}}{2\left(-50\right)}$$

Add 250000 to -1440000.

$$x=\frac{-500±\sqrt{-1190000}}{2\left(-50\right)}$$

Take the square root of -1190000.

$$x=\frac{-500±100\sqrt{119}i}{2\left(-50\right)}$$

Multiply 2 times -50.

$$x=\frac{-500±100\sqrt{119}i}{-100}$$

Now solve the equation x=\frac{-500±100\sqrt{119}i}{-100} when ± is plus. Add -500 to 100i\sqrt{119}.

$$x=\frac{-500+100\sqrt{119}i}{-100}$$

Divide -500+100i\sqrt{119} by -100.

$$x=-\sqrt{119}i+5$$

Now solve the equation x=\frac{-500±100\sqrt{119}i}{-100} when ± is minus. Subtract 100i\sqrt{119} from -500.

$$x=\frac{-100\sqrt{119}i-500}{-100}$$

Divide -500-100i\sqrt{119} by -100.

$$x=5+\sqrt{119}i$$

The equation is now solved.

$$x=-\sqrt{119}i+5 x=5+\sqrt{119}i$$

Steps for Completing the Square

Variable x cannot be equal to any of the values 0,10 since division by zero is not defined. Multiply both sides of the equation by x\left(x-10\right), the least common multiple of x,x-10.

$$\left(x-10\right)\times 720=x\times 720+x\left(x-10\right)\times 50$$

Use the distributive property to multiply x-10 by 720.

$$720x-7200=x\times 720+x\left(x-10\right)\times 50$$

Use the distributive property to multiply x by x-10.

$$720x-7200=x\times 720+\left(x^{2}-10x\right)\times 50$$

Use the distributive property to multiply x^{2}-10x by 50.

$$720x-7200=x\times 720+50x^{2}-500x$$

Combine x\times 720 and -500x to get 220x.

$$720x-7200=220x+50x^{2}$$

Subtract 220x from both sides.

$$720x-7200-220x=50x^{2}$$

Combine 720x and -220x to get 500x.

$$500x-7200=50x^{2}$$

Subtract 50x^{2} from both sides.

$$500x-7200-50x^{2}=0$$

Add 7200 to both sides. Anything plus zero gives itself.

$$500x-50x^{2}=7200$$

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.

$$-50x^{2}+500x=7200$$

Divide both sides by -50.

$$\frac{-50x^{2}+500x}{-50}=\frac{7200}{-50}$$

Dividing by -50 undoes the multiplication by -50.

$$x^{2}+\frac{500}{-50}x=\frac{7200}{-50}$$

Divide 500 by -50.

$$x^{2}-10x=\frac{7200}{-50}$$

Divide 7200 by -50.

$$x^{2}-10x=-144$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-144+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-144+25$$

Add -144 to 25.

$$x^{2}-10x+25=-119$$

Factor x^{2}-10x+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}.

$$\left(x-5\right)^{2}=-119$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{-119}$$

Simplify.

$$x-5=\sqrt{119}i x-5=-\sqrt{119}i$$

Add 5 to both sides of the equation.

$$x=5+\sqrt{119}i x=-\sqrt{119}i+5$$