Solve 2x+400/x*240=6x-11

$$\frac { 2 x + 400 } { x \cdot 240 } = 6 x - 11$$

Solve for x

$$x = \frac{\sqrt{2321041} + 1321}{1440} \approx 1.975344658$$

$$x=\frac{1321-\sqrt{2321041}}{1440}\approx -0.140622436$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 240x.

$$2x+400=6x\times 240x+240x\left(-11\right)$$

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

$$2x+400=6x^{2}\times 240+240x\left(-11\right)$$

Multiply 6 and 240 to get 1440.

$$2x+400=1440x^{2}+240x\left(-11\right)$$

Multiply 240 and -11 to get -2640.

$$2x+400=1440x^{2}-2640x$$

Subtract 1440x^{2} from both sides.

$$2x+400-1440x^{2}=-2640x$$

Add 2640x to both sides.

$$2x+400-1440x^{2}+2640x=0$$

Combine 2x and 2640x to get 2642x.

$$2642x+400-1440x^{2}=0$$

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.

$$-1440x^{2}+2642x+400=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1440 for a, 2642 for b, and 400 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2642±\sqrt{2642^{2}-4\left(-1440\right)\times 400}}{2\left(-1440\right)}$$

Square 2642.

$$x=\frac{-2642±\sqrt{6980164-4\left(-1440\right)\times 400}}{2\left(-1440\right)}$$

Multiply -4 times -1440.

$$x=\frac{-2642±\sqrt{6980164+5760\times 400}}{2\left(-1440\right)}$$

Multiply 5760 times 400.

$$x=\frac{-2642±\sqrt{6980164+2304000}}{2\left(-1440\right)}$$

Add 6980164 to 2304000.

$$x=\frac{-2642±\sqrt{9284164}}{2\left(-1440\right)}$$

Take the square root of 9284164.

$$x=\frac{-2642±2\sqrt{2321041}}{2\left(-1440\right)}$$

Multiply 2 times -1440.

$$x=\frac{-2642±2\sqrt{2321041}}{-2880}$$

Now solve the equation x=\frac{-2642±2\sqrt{2321041}}{-2880} when ± is plus. Add -2642 to 2\sqrt{2321041}.

$$x=\frac{2\sqrt{2321041}-2642}{-2880}$$

Divide -2642+2\sqrt{2321041} by -2880.

$$x=\frac{1321-\sqrt{2321041}}{1440}$$

Now solve the equation x=\frac{-2642±2\sqrt{2321041}}{-2880} when ± is minus. Subtract 2\sqrt{2321041} from -2642.

$$x=\frac{-2\sqrt{2321041}-2642}{-2880}$$

Divide -2642-2\sqrt{2321041} by -2880.

$$x=\frac{\sqrt{2321041}+1321}{1440}$$

The equation is now solved.

$$x=\frac{1321-\sqrt{2321041}}{1440} x=\frac{\sqrt{2321041}+1321}{1440}$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 240x.

$$2x+400=6x\times 240x+240x\left(-11\right)$$

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

$$2x+400=6x^{2}\times 240+240x\left(-11\right)$$

Multiply 6 and 240 to get 1440.

$$2x+400=1440x^{2}+240x\left(-11\right)$$

Multiply 240 and -11 to get -2640.

$$2x+400=1440x^{2}-2640x$$

Subtract 1440x^{2} from both sides.

$$2x+400-1440x^{2}=-2640x$$

Add 2640x to both sides.

$$2x+400-1440x^{2}+2640x=0$$

Combine 2x and 2640x to get 2642x.

$$2642x+400-1440x^{2}=0$$

Subtract 400 from both sides. Anything subtracted from zero gives its negation.

$$2642x-1440x^{2}=-400$$

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.

$$-1440x^{2}+2642x=-400$$

Divide both sides by -1440.

$$\frac{-1440x^{2}+2642x}{-1440}=-\frac{400}{-1440}$$

Dividing by -1440 undoes the multiplication by -1440.

$$x^{2}+\frac{2642}{-1440}x=-\frac{400}{-1440}$$

Reduce the fraction \frac{2642}{-1440} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1321}{720}x=-\frac{400}{-1440}$$

Reduce the fraction \frac{-400}{-1440} to lowest terms by extracting and canceling out 80.

$$x^{2}-\frac{1321}{720}x=\frac{5}{18}$$

Divide -\frac{1321}{720}, the coefficient of the x term, by 2 to get -\frac{1321}{1440}. Then add the square of -\frac{1321}{1440} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1321}{720}x+\left(-\frac{1321}{1440}\right)^{2}=\frac{5}{18}+\left(-\frac{1321}{1440}\right)^{2}$$

Square -\frac{1321}{1440} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1321}{720}x+\frac{1745041}{2073600}=\frac{5}{18}+\frac{1745041}{2073600}$$

Add \frac{5}{18} to \frac{1745041}{2073600} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1321}{720}x+\frac{1745041}{2073600}=\frac{2321041}{2073600}$$

Factor x^{2}-\frac{1321}{720}x+\frac{1745041}{2073600}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1321}{1440}\right)^{2}=\frac{2321041}{2073600}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1321}{1440}\right)^{2}}=\sqrt{\frac{2321041}{2073600}}$$

Simplify.

$$x-\frac{1321}{1440}=\frac{\sqrt{2321041}}{1440} x-\frac{1321}{1440}=-\frac{\sqrt{2321041}}{1440}$$

Add \frac{1321}{1440} to both sides of the equation.

$$x=\frac{\sqrt{2321041}+1321}{1440} x=\frac{1321-\sqrt{2321041}}{1440}$$