Solve 2x(x/3+1/4)-(x+1)(3-2x)/3=1/2*(frac{2}{3}-frac{2}{5}x)-frac{5}{2}

$$2 x ( \frac { x } { 3 } + \frac { 1 } { 4 } ) - \frac { ( x + 1 ) ( 3 - 2 x ) } { 3 } = \frac { 1 } { 2 } \times ( \frac { 2 } { 3 } - \frac { 2 } { 5 } x ) - \frac { 5 } { 2 }$$

Solve for x (complex solution)

$$x=\frac{-11+\sqrt{5479}i}{80}\approx -0.1375+0.925253344i$$

$$x=\frac{-\sqrt{5479}i-11}{80}\approx -0.1375-0.925253344i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 60, the least common multiple of 3,4,2,5.

$$120x\left(\frac{x}{3}+\frac{1}{4}\right)-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 4 is 12. Multiply \frac{x}{3} times \frac{4}{4}. Multiply \frac{1}{4} times \frac{3}{3}.

$$120x\left(\frac{4x}{12}+\frac{3}{12}\right)-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Since \frac{4x}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$120x\times \frac{4x+3}{12}-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Cancel out 12, the greatest common factor in 120 and 12.

$$10\left(4x+3\right)x-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Use the distributive property to multiply -20 by x+1.

$$10\left(4x+3\right)x+\left(-20x-20\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Use the distributive property to multiply -20x-20 by 3-2x and combine like terms.

$$10\left(4x+3\right)x-20x+40x^{2}-60=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Use the distributive property to multiply 30 by \frac{2}{3}-\frac{2}{5}x.

$$10\left(4x+3\right)x-20x+40x^{2}-60=20-12x-150$$

Subtract 150 from 20 to get -130.

$$10\left(4x+3\right)x-20x+40x^{2}-60=-130-12x$$

Use the distributive property to multiply 10 by 4x+3.

$$\left(40x+30\right)x-20x+40x^{2}-60=-130-12x$$

Use the distributive property to multiply 40x+30 by x.

$$40x^{2}+30x-20x+40x^{2}-60=-130-12x$$

Combine 30x and -20x to get 10x.

$$40x^{2}+10x+40x^{2}-60=-130-12x$$

Combine 40x^{2} and 40x^{2} to get 80x^{2}.

$$80x^{2}+10x-60=-130-12x$$

Subtract -130 from both sides.

$$80x^{2}+10x-60-\left(-130\right)=-12x$$

The opposite of -130 is 130.

$$80x^{2}+10x-60+130=-12x$$

Add 12x to both sides.

$$80x^{2}+10x-60+130+12x=0$$

Add -60 and 130 to get 70.

$$80x^{2}+10x+70+12x=0$$

Combine 10x and 12x to get 22x.

$$80x^{2}+22x+70=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 80 for a, 22 for b, and 70 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-22±\sqrt{22^{2}-4\times 80\times 70}}{2\times 80}$$

Square 22.

$$x=\frac{-22±\sqrt{484-4\times 80\times 70}}{2\times 80}$$

Multiply -4 times 80.

$$x=\frac{-22±\sqrt{484-320\times 70}}{2\times 80}$$

Multiply -320 times 70.

$$x=\frac{-22±\sqrt{484-22400}}{2\times 80}$$

Add 484 to -22400.

$$x=\frac{-22±\sqrt{-21916}}{2\times 80}$$

Take the square root of -21916.

$$x=\frac{-22±2\sqrt{5479}i}{2\times 80}$$

Multiply 2 times 80.

$$x=\frac{-22±2\sqrt{5479}i}{160}$$

Now solve the equation x=\frac{-22±2\sqrt{5479}i}{160} when ± is plus. Add -22 to 2i\sqrt{5479}.

$$x=\frac{-22+2\sqrt{5479}i}{160}$$

Divide -22+2i\sqrt{5479} by 160.

$$x=\frac{-11+\sqrt{5479}i}{80}$$

Now solve the equation x=\frac{-22±2\sqrt{5479}i}{160} when ± is minus. Subtract 2i\sqrt{5479} from -22.

$$x=\frac{-2\sqrt{5479}i-22}{160}$$

Divide -22-2i\sqrt{5479} by 160.

$$x=\frac{-\sqrt{5479}i-11}{80}$$

The equation is now solved.

$$x=\frac{-11+\sqrt{5479}i}{80} x=\frac{-\sqrt{5479}i-11}{80}$$

Steps for Completing the Square

Multiply both sides of the equation by 60, the least common multiple of 3,4,2,5.

$$120x\left(\frac{x}{3}+\frac{1}{4}\right)-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 3 and 4 is 12. Multiply \frac{x}{3} times \frac{4}{4}. Multiply \frac{1}{4} times \frac{3}{3}.

$$120x\left(\frac{4x}{12}+\frac{3}{12}\right)-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Since \frac{4x}{12} and \frac{3}{12} have the same denominator, add them by adding their numerators.

$$120x\times \frac{4x+3}{12}-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Cancel out 12, the greatest common factor in 120 and 12.

$$10\left(4x+3\right)x-20\left(x+1\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Use the distributive property to multiply -20 by x+1.

$$10\left(4x+3\right)x+\left(-20x-20\right)\left(3-2x\right)=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Use the distributive property to multiply -20x-20 by 3-2x and combine like terms.

$$10\left(4x+3\right)x-20x+40x^{2}-60=30\left(\frac{2}{3}-\frac{2}{5}x\right)-150$$

Use the distributive property to multiply 30 by \frac{2}{3}-\frac{2}{5}x.

$$10\left(4x+3\right)x-20x+40x^{2}-60=20-12x-150$$

Subtract 150 from 20 to get -130.

$$10\left(4x+3\right)x-20x+40x^{2}-60=-130-12x$$

Use the distributive property to multiply 10 by 4x+3.

$$\left(40x+30\right)x-20x+40x^{2}-60=-130-12x$$

Use the distributive property to multiply 40x+30 by x.

$$40x^{2}+30x-20x+40x^{2}-60=-130-12x$$

Combine 30x and -20x to get 10x.

$$40x^{2}+10x+40x^{2}-60=-130-12x$$

Combine 40x^{2} and 40x^{2} to get 80x^{2}.

$$80x^{2}+10x-60=-130-12x$$

Add 12x to both sides.

$$80x^{2}+10x-60+12x=-130$$

Combine 10x and 12x to get 22x.

$$80x^{2}+22x-60=-130$$

Add 60 to both sides.

$$80x^{2}+22x=-130+60$$

Add -130 and 60 to get -70.

$$80x^{2}+22x=-70$$

Divide both sides by 80.

$$\frac{80x^{2}+22x}{80}=-\frac{70}{80}$$

Dividing by 80 undoes the multiplication by 80.

$$x^{2}+\frac{22}{80}x=-\frac{70}{80}$$

Reduce the fraction \frac{22}{80} to lowest terms by extracting and canceling out 2.

$$x^{2}+\frac{11}{40}x=-\frac{70}{80}$$

Reduce the fraction \frac{-70}{80} to lowest terms by extracting and canceling out 10.

$$x^{2}+\frac{11}{40}x=-\frac{7}{8}$$

Divide \frac{11}{40}, the coefficient of the x term, by 2 to get \frac{11}{80}. Then add the square of \frac{11}{80} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{40}x+\left(\frac{11}{80}\right)^{2}=-\frac{7}{8}+\left(\frac{11}{80}\right)^{2}$$

Square \frac{11}{80} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{40}x+\frac{121}{6400}=-\frac{7}{8}+\frac{121}{6400}$$

Add -\frac{7}{8} to \frac{121}{6400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{11}{40}x+\frac{121}{6400}=-\frac{5479}{6400}$$

Factor x^{2}+\frac{11}{40}x+\frac{121}{6400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{80}\right)^{2}=-\frac{5479}{6400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{80}\right)^{2}}=\sqrt{-\frac{5479}{6400}}$$

Simplify.

$$x+\frac{11}{80}=\frac{\sqrt{5479}i}{80} x+\frac{11}{80}=-\frac{\sqrt{5479}i}{80}$$

Subtract \frac{11}{80} from both sides of the equation.

$$x=\frac{-11+\sqrt{5479}i}{80} x=\frac{-\sqrt{5479}i-11}{80}$$