Solve 180/6x-180/(6+2x)=3

$$\frac { 180 } { 6 x } - \frac { 180 } { ( 6 + 2 x ) } = 3$$

Solve for x

$$x = \frac{\sqrt{649} - 23}{2} \approx 1.237739203$$

$$x=\frac{-\sqrt{649}-23}{2}\approx -24.237739203$$

Steps Using the Quadratic Formula

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 6x\left(x+3\right), the least common multiple of 6x,6+2x.

$$\left(x+3\right)\times 180-3x\times 180=18x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 180.

$$180x+540-3x\times 180=18x\left(x+3\right)$$

Multiply 3 and 180 to get 540.

$$180x+540-540x=18x\left(x+3\right)$$

Use the distributive property to multiply 18x by x+3.

$$180x+540-540x=18x^{2}+54x$$

Subtract 18x^{2} from both sides.

$$180x+540-540x-18x^{2}=54x$$

Subtract 54x from both sides.

$$180x+540-540x-18x^{2}-54x=0$$

Combine 180x and -54x to get 126x.

$$126x+540-540x-18x^{2}=0$$

Combine 126x and -540x to get -414x.

$$-414x+540-18x^{2}=0$$

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.

$$-18x^{2}-414x+540=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, -414 for b, and 540 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-414\right)±\sqrt{\left(-414\right)^{2}-4\left(-18\right)\times 540}}{2\left(-18\right)}$$

Square -414.

$$x=\frac{-\left(-414\right)±\sqrt{171396-4\left(-18\right)\times 540}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-\left(-414\right)±\sqrt{171396+72\times 540}}{2\left(-18\right)}$$

Multiply 72 times 540.

$$x=\frac{-\left(-414\right)±\sqrt{171396+38880}}{2\left(-18\right)}$$

Add 171396 to 38880.

$$x=\frac{-\left(-414\right)±\sqrt{210276}}{2\left(-18\right)}$$

Take the square root of 210276.

$$x=\frac{-\left(-414\right)±18\sqrt{649}}{2\left(-18\right)}$$

The opposite of -414 is 414.

$$x=\frac{414±18\sqrt{649}}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{414±18\sqrt{649}}{-36}$$

Now solve the equation x=\frac{414±18\sqrt{649}}{-36} when ± is plus. Add 414 to 18\sqrt{649}.

$$x=\frac{18\sqrt{649}+414}{-36}$$

Divide 414+18\sqrt{649} by -36.

$$x=\frac{-\sqrt{649}-23}{2}$$

Now solve the equation x=\frac{414±18\sqrt{649}}{-36} when ± is minus. Subtract 18\sqrt{649} from 414.

$$x=\frac{414-18\sqrt{649}}{-36}$$

Divide 414-18\sqrt{649} by -36.

$$x=\frac{\sqrt{649}-23}{2}$$

The equation is now solved.

$$x=\frac{-\sqrt{649}-23}{2} x=\frac{\sqrt{649}-23}{2}$$

Steps for Completing the Square

Variable x cannot be equal to any of the values -3,0 since division by zero is not defined. Multiply both sides of the equation by 6x\left(x+3\right), the least common multiple of 6x,6+2x.

$$\left(x+3\right)\times 180-3x\times 180=18x\left(x+3\right)$$

Use the distributive property to multiply x+3 by 180.

$$180x+540-3x\times 180=18x\left(x+3\right)$$

Multiply 3 and 180 to get 540.

$$180x+540-540x=18x\left(x+3\right)$$

Use the distributive property to multiply 18x by x+3.

$$180x+540-540x=18x^{2}+54x$$

Subtract 18x^{2} from both sides.

$$180x+540-540x-18x^{2}=54x$$

Subtract 54x from both sides.

$$180x+540-540x-18x^{2}-54x=0$$

Combine 180x and -54x to get 126x.

$$126x+540-540x-18x^{2}=0$$

Subtract 540 from both sides. Anything subtracted from zero gives its negation.

$$126x-540x-18x^{2}=-540$$

Combine 126x and -540x to get -414x.

$$-414x-18x^{2}=-540$$

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.

$$-18x^{2}-414x=-540$$

Divide both sides by -18.

$$\frac{-18x^{2}-414x}{-18}=-\frac{540}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\left(-\frac{414}{-18}\right)x=-\frac{540}{-18}$$

Divide -414 by -18.

$$x^{2}+23x=-\frac{540}{-18}$$

Divide -540 by -18.

$$x^{2}+23x=30$$

Divide 23, the coefficient of the x term, by 2 to get \frac{23}{2}. Then add the square of \frac{23}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+23x+\left(\frac{23}{2}\right)^{2}=30+\left(\frac{23}{2}\right)^{2}$$

Square \frac{23}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+23x+\frac{529}{4}=30+\frac{529}{4}$$

Add 30 to \frac{529}{4}.

$$x^{2}+23x+\frac{529}{4}=\frac{649}{4}$$

Factor x^{2}+23x+\frac{529}{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}.

$$\left(x+\frac{23}{2}\right)^{2}=\frac{649}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{23}{2}\right)^{2}}=\sqrt{\frac{649}{4}}$$

Simplify.

$$x+\frac{23}{2}=\frac{\sqrt{649}}{2} x+\frac{23}{2}=-\frac{\sqrt{649}}{2}$$

Subtract \frac{23}{2} from both sides of the equation.

$$x=\frac{\sqrt{649}-23}{2} x=\frac{-\sqrt{649}-23}{2}$$