Solve 5(1+x)(1-2x)/20*frac{5{2}-5(1-x)}=15/60

$$\frac { 5 ( 1 + x ) ( 1 - 2 x ) } { 20 \times \frac { 5 } { 2 } - 5 ( 1 - x ) } = \frac { 15 } { 60 }$$

Solve for x (complex solution)

$$x=\frac{-3\sqrt{15}i-5}{16}\approx -0.3125-0.726184377i$$

$$x=\frac{-5+3\sqrt{15}i}{16}\approx -0.3125+0.726184377i$$

Steps Using the Quadratic Formula

Multiply 20 and \frac{5}{2} to get 50.

$$\frac{5\left(1+x\right)\left(1-2x\right)}{50-5\left(1-x\right)}=\frac{15}{60}$$

Factor the expressions that are not already factored in \frac{5\left(1+x\right)\left(1-2x\right)}{50-5\left(1-x\right)}.

$$\frac{5\left(x+1\right)\left(-2x+1\right)}{5\left(x+9\right)}=\frac{15}{60}$$

Cancel out 5 in both numerator and denominator.

$$\frac{\left(x+1\right)\left(-2x+1\right)}{x+9}=\frac{15}{60}$$

Reduce the fraction \frac{15}{60} to lowest terms by extracting and canceling out 15.

$$\frac{\left(x+1\right)\left(-2x+1\right)}{x+9}=\frac{1}{4}$$

Use the distributive property to multiply x+1 by -2x+1 and combine like terms.

$$\frac{-2x^{2}-x+1}{x+9}=\frac{1}{4}$$

Subtract \frac{1}{4} from both sides.

$$\frac{-2x^{2}-x+1}{x+9}-\frac{1}{4}=0$$

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of x+9 and 4 is 4\left(x+9\right). Multiply \frac{-2x^{2}-x+1}{x+9} times \frac{4}{4}. Multiply \frac{1}{4} times \frac{x+9}{x+9}.

$$\frac{4\left(-2x^{2}-x+1\right)}{4\left(x+9\right)}-\frac{x+9}{4\left(x+9\right)}=0$$

Since \frac{4\left(-2x^{2}-x+1\right)}{4\left(x+9\right)} and \frac{x+9}{4\left(x+9\right)} have the same denominator, subtract them by subtracting their numerators.

$$\frac{4\left(-2x^{2}-x+1\right)-\left(x+9\right)}{4\left(x+9\right)}=0$$

Do the multiplications in 4\left(-2x^{2}-x+1\right)-\left(x+9\right).

$$\frac{-8x^{2}-4x+4-x-9}{4\left(x+9\right)}=0$$

Combine like terms in -8x^{2}-4x+4-x-9.

$$\frac{-8x^{2}-5x-5}{4\left(x+9\right)}=0$$

Variable x cannot be equal to -9 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+9\right).

$$-8x^{2}-5x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, -5 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\left(-8\right)\left(-5\right)}}{2\left(-8\right)}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\left(-8\right)\left(-5\right)}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$x=\frac{-\left(-5\right)±\sqrt{25+32\left(-5\right)}}{2\left(-8\right)}$$

Multiply 32 times -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-160}}{2\left(-8\right)}$$

Add 25 to -160.

$$x=\frac{-\left(-5\right)±\sqrt{-135}}{2\left(-8\right)}$$

Take the square root of -135.

$$x=\frac{-\left(-5\right)±3\sqrt{15}i}{2\left(-8\right)}$$

The opposite of -5 is 5.

$$x=\frac{5±3\sqrt{15}i}{2\left(-8\right)}$$

Multiply 2 times -8.

$$x=\frac{5±3\sqrt{15}i}{-16}$$

Now solve the equation x=\frac{5±3\sqrt{15}i}{-16} when ± is plus. Add 5 to 3i\sqrt{15}.

$$x=\frac{5+3\sqrt{15}i}{-16}$$

Divide 5+3i\sqrt{15} by -16.

$$x=\frac{-3\sqrt{15}i-5}{16}$$

Now solve the equation x=\frac{5±3\sqrt{15}i}{-16} when ± is minus. Subtract 3i\sqrt{15} from 5.

$$x=\frac{-3\sqrt{15}i+5}{-16}$$

Divide 5-3i\sqrt{15} by -16.

$$x=\frac{-5+3\sqrt{15}i}{16}$$

The equation is now solved.

$$x=\frac{-3\sqrt{15}i-5}{16} x=\frac{-5+3\sqrt{15}i}{16}$$

Steps for Completing the Square

Multiply 20 and \frac{5}{2} to get 50.

$$\frac{5\left(1+x\right)\left(1-2x\right)}{50-5\left(1-x\right)}=\frac{15}{60}$$

Factor the expressions that are not already factored in \frac{5\left(1+x\right)\left(1-2x\right)}{50-5\left(1-x\right)}.

$$\frac{5\left(x+1\right)\left(-2x+1\right)}{5\left(x+9\right)}=\frac{15}{60}$$

Cancel out 5 in both numerator and denominator.

$$\frac{\left(x+1\right)\left(-2x+1\right)}{x+9}=\frac{15}{60}$$

Reduce the fraction \frac{15}{60} to lowest terms by extracting and canceling out 15.

$$\frac{\left(x+1\right)\left(-2x+1\right)}{x+9}=\frac{1}{4}$$

Use the distributive property to multiply x+1 by -2x+1 and combine like terms.

$$\frac{-2x^{2}-x+1}{x+9}=\frac{1}{4}$$

Variable x cannot be equal to -9 since division by zero is not defined. Multiply both sides of the equation by 4\left(x+9\right), the least common multiple of x+9,4.

$$4\left(-2x^{2}-x+1\right)=x+9$$

Use the distributive property to multiply 4 by -2x^{2}-x+1.

$$-8x^{2}-4x+4=x+9$$

Subtract x from both sides.

$$-8x^{2}-4x+4-x=9$$

Combine -4x and -x to get -5x.

$$-8x^{2}-5x+4=9$$

Subtract 4 from both sides.

$$-8x^{2}-5x=9-4$$

Subtract 4 from 9 to get 5.

$$-8x^{2}-5x=5$$

Divide both sides by -8.

$$\frac{-8x^{2}-5x}{-8}=\frac{5}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$x^{2}+\left(-\frac{5}{-8}\right)x=\frac{5}{-8}$$

Divide -5 by -8.

$$x^{2}+\frac{5}{8}x=\frac{5}{-8}$$

Divide 5 by -8.

$$x^{2}+\frac{5}{8}x=-\frac{5}{8}$$

Divide \frac{5}{8}, the coefficient of the x term, by 2 to get \frac{5}{16}. Then add the square of \frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5}{8}x+\left(\frac{5}{16}\right)^{2}=-\frac{5}{8}+\left(\frac{5}{16}\right)^{2}$$

Square \frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=-\frac{5}{8}+\frac{25}{256}$$

Add -\frac{5}{8} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5}{8}x+\frac{25}{256}=-\frac{135}{256}$$

Factor x^{2}+\frac{5}{8}x+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5}{16}\right)^{2}=-\frac{135}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{16}\right)^{2}}=\sqrt{-\frac{135}{256}}$$

Simplify.

$$x+\frac{5}{16}=\frac{3\sqrt{15}i}{16} x+\frac{5}{16}=-\frac{3\sqrt{15}i}{16}$$

Subtract \frac{5}{16} from both sides of the equation.

$$x=\frac{-5+3\sqrt{15}i}{16} x=\frac{-3\sqrt{15}i-5}{16}$$