Solve 32/3x1/6-3/4left(2x+18right)=-4

$$3 \frac{ 2 }{ 3x } \frac{ 1 }{ 6 } - \frac{ 3 }{ 4 } \left( 2x+18 \right) = -4$$

Solve for x

$$x=\frac{\sqrt{41}}{2}-\frac{19}{6}\approx 0.034895452$$

$$x=-\frac{\sqrt{41}}{2}-\frac{19}{6}\approx -6.368228785$$

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 12x, the least common multiple of 3x,6,4.

$$3\times 4\times 2\times \frac{1}{6}-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply 3 and 4 to get 12.

$$12\times 2\times \frac{1}{6}-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply 12 and 2 to get 24.

$$24\times \frac{1}{6}-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply 24 and \frac{1}{6} to get 4.

$$4-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply -\frac{3}{4} and 12 to get -9.

$$4-9\left(2x+18\right)x=-48x$$

Use the distributive property to multiply -9 by 2x+18.

$$4+\left(-18x-162\right)x=-48x$$

Use the distributive property to multiply -18x-162 by x.

$$4-18x^{2}-162x=-48x$$

Add 48x to both sides.

$$4-18x^{2}-162x+48x=0$$

Combine -162x and 48x to get -114x.

$$4-18x^{2}-114x=0$$

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.

$$-18x^{2}-114x+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -18 for a, -114 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-114\right)±\sqrt{\left(-114\right)^{2}-4\left(-18\right)\times 4}}{2\left(-18\right)}$$

Square -114.

$$x=\frac{-\left(-114\right)±\sqrt{12996-4\left(-18\right)\times 4}}{2\left(-18\right)}$$

Multiply -4 times -18.

$$x=\frac{-\left(-114\right)±\sqrt{12996+72\times 4}}{2\left(-18\right)}$$

Multiply 72 times 4.

$$x=\frac{-\left(-114\right)±\sqrt{12996+288}}{2\left(-18\right)}$$

Add 12996 to 288.

$$x=\frac{-\left(-114\right)±\sqrt{13284}}{2\left(-18\right)}$$

Take the square root of 13284.

$$x=\frac{-\left(-114\right)±18\sqrt{41}}{2\left(-18\right)}$$

The opposite of -114 is 114.

$$x=\frac{114±18\sqrt{41}}{2\left(-18\right)}$$

Multiply 2 times -18.

$$x=\frac{114±18\sqrt{41}}{-36}$$

Now solve the equation x=\frac{114±18\sqrt{41}}{-36} when ± is plus. Add 114 to 18\sqrt{41}.

$$x=\frac{18\sqrt{41}+114}{-36}$$

Divide 114+18\sqrt{41} by -36.

$$x=-\frac{\sqrt{41}}{2}-\frac{19}{6}$$

Now solve the equation x=\frac{114±18\sqrt{41}}{-36} when ± is minus. Subtract 18\sqrt{41} from 114.

$$x=\frac{114-18\sqrt{41}}{-36}$$

Divide 114-18\sqrt{41} by -36.

$$x=\frac{\sqrt{41}}{2}-\frac{19}{6}$$

The equation is now solved.

$$x=-\frac{\sqrt{41}}{2}-\frac{19}{6} x=\frac{\sqrt{41}}{2}-\frac{19}{6}$$

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 12x, the least common multiple of 3x,6,4.

$$3\times 4\times 2\times \frac{1}{6}-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply 3 and 4 to get 12.

$$12\times 2\times \frac{1}{6}-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply 12 and 2 to get 24.

$$24\times \frac{1}{6}-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply 24 and \frac{1}{6} to get 4.

$$4-\frac{3}{4}\left(2x+18\right)\times 12x=-48x$$

Multiply -\frac{3}{4} and 12 to get -9.

$$4-9\left(2x+18\right)x=-48x$$

Use the distributive property to multiply -9 by 2x+18.

$$4+\left(-18x-162\right)x=-48x$$

Use the distributive property to multiply -18x-162 by x.

$$4-18x^{2}-162x=-48x$$

Add 48x to both sides.

$$4-18x^{2}-162x+48x=0$$

Combine -162x and 48x to get -114x.

$$4-18x^{2}-114x=0$$

Subtract 4 from both sides. Anything subtracted from zero gives its negation.

$$-18x^{2}-114x=-4$$

Divide both sides by -18.

$$\frac{-18x^{2}-114x}{-18}=-\frac{4}{-18}$$

Dividing by -18 undoes the multiplication by -18.

$$x^{2}+\left(-\frac{114}{-18}\right)x=-\frac{4}{-18}$$

Reduce the fraction \frac{-114}{-18} to lowest terms by extracting and canceling out 6.

$$x^{2}+\frac{19}{3}x=-\frac{4}{-18}$$

Reduce the fraction \frac{-4}{-18} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{19}{3}x=\frac{2}{9}$$

Divide \frac{19}{3}, the coefficient of the x term, by 2 to get \frac{19}{6}. Then add the square of \frac{19}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{19}{3}x+\left(\frac{19}{6}\right)^{2}=\frac{2}{9}+\left(\frac{19}{6}\right)^{2}$$

Square \frac{19}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{19}{3}x+\frac{361}{36}=\frac{2}{9}+\frac{361}{36}$$

Add \frac{2}{9} to \frac{361}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{19}{3}x+\frac{361}{36}=\frac{41}{4}$$

Factor x^{2}+\frac{19}{3}x+\frac{361}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{19}{6}\right)^{2}=\frac{41}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{19}{6}\right)^{2}}=\sqrt{\frac{41}{4}}$$

Simplify.

$$x+\frac{19}{6}=\frac{\sqrt{41}}{2} x+\frac{19}{6}=-\frac{\sqrt{41}}{2}$$

Subtract \frac{19}{6} from both sides of the equation.

$$x=\frac{\sqrt{41}}{2}-\frac{19}{6} x=-\frac{\sqrt{41}}{2}-\frac{19}{6}$$