Solve 1/5x-3=5x1/10(x+1)

$$\frac { 1 } { 5 } x - 3 = 5 x \frac { 1 } { 10 } ( x + 1 )$$

Solve for x (complex solution)

$$x=\frac{-3+\sqrt{591}i}{10}\approx -0.3+2.431049156i$$

$$x=\frac{-\sqrt{591}i-3}{10}\approx -0.3-2.431049156i$$

Steps Using the Quadratic Formula

Multiply 5 and \frac{1}{10} to get \frac{5}{10}.

$$\frac{1}{5}x-3=\frac{5}{10}x\left(x+1\right)$$

Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$\frac{1}{5}x-3=\frac{1}{2}x\left(x+1\right)$$

Use the distributive property to multiply \frac{1}{2}x by x+1.

$$\frac{1}{5}x-3=\frac{1}{2}xx+\frac{1}{2}x$$

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

$$\frac{1}{5}x-3=\frac{1}{2}x^{2}+\frac{1}{2}x$$

Subtract \frac{1}{2}x^{2} from both sides.

$$\frac{1}{5}x-3-\frac{1}{2}x^{2}=\frac{1}{2}x$$

Subtract \frac{1}{2}x from both sides.

$$\frac{1}{5}x-3-\frac{1}{2}x^{2}-\frac{1}{2}x=0$$

Combine \frac{1}{5}x and -\frac{1}{2}x to get -\frac{3}{10}x.

$$-\frac{3}{10}x-3-\frac{1}{2}x^{2}=0$$

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.

$$-\frac{1}{2}x^{2}-\frac{3}{10}x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -\frac{1}{2} for a, -\frac{3}{10} for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-\frac{3}{10}\right)±\sqrt{\left(-\frac{3}{10}\right)^{2}-4\left(-\frac{1}{2}\right)\left(-3\right)}}{2\left(-\frac{1}{2}\right)}$$

Square -\frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x=\frac{-\left(-\frac{3}{10}\right)±\sqrt{\frac{9}{100}-4\left(-\frac{1}{2}\right)\left(-3\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply -4 times -\frac{1}{2}.

$$x=\frac{-\left(-\frac{3}{10}\right)±\sqrt{\frac{9}{100}+2\left(-3\right)}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -3.

$$x=\frac{-\left(-\frac{3}{10}\right)±\sqrt{\frac{9}{100}-6}}{2\left(-\frac{1}{2}\right)}$$

Add \frac{9}{100} to -6.

$$x=\frac{-\left(-\frac{3}{10}\right)±\sqrt{-\frac{591}{100}}}{2\left(-\frac{1}{2}\right)}$$

Take the square root of -\frac{591}{100}.

$$x=\frac{-\left(-\frac{3}{10}\right)±\frac{\sqrt{591}i}{10}}{2\left(-\frac{1}{2}\right)}$$

The opposite of -\frac{3}{10} is \frac{3}{10}.

$$x=\frac{\frac{3}{10}±\frac{\sqrt{591}i}{10}}{2\left(-\frac{1}{2}\right)}$$

Multiply 2 times -\frac{1}{2}.

$$x=\frac{\frac{3}{10}±\frac{\sqrt{591}i}{10}}{-1}$$

Now solve the equation x=\frac{\frac{3}{10}±\frac{\sqrt{591}i}{10}}{-1} when ± is plus. Add \frac{3}{10} to \frac{i\sqrt{591}}{10}.

$$x=\frac{3+\sqrt{591}i}{-10}$$

Divide \frac{3+i\sqrt{591}}{10} by -1.

$$x=\frac{-\sqrt{591}i-3}{10}$$

Now solve the equation x=\frac{\frac{3}{10}±\frac{\sqrt{591}i}{10}}{-1} when ± is minus. Subtract \frac{i\sqrt{591}}{10} from \frac{3}{10}.

$$x=\frac{-\sqrt{591}i+3}{-10}$$

Divide \frac{3-i\sqrt{591}}{10} by -1.

$$x=\frac{-3+\sqrt{591}i}{10}$$

The equation is now solved.

$$x=\frac{-\sqrt{591}i-3}{10} x=\frac{-3+\sqrt{591}i}{10}$$

Steps for Completing the Square

Multiply 5 and \frac{1}{10} to get \frac{5}{10}.

$$\frac{1}{5}x-3=\frac{5}{10}x\left(x+1\right)$$

Reduce the fraction \frac{5}{10} to lowest terms by extracting and canceling out 5.

$$\frac{1}{5}x-3=\frac{1}{2}x\left(x+1\right)$$

Use the distributive property to multiply \frac{1}{2}x by x+1.

$$\frac{1}{5}x-3=\frac{1}{2}xx+\frac{1}{2}x$$

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

$$\frac{1}{5}x-3=\frac{1}{2}x^{2}+\frac{1}{2}x$$

Subtract \frac{1}{2}x^{2} from both sides.

$$\frac{1}{5}x-3-\frac{1}{2}x^{2}=\frac{1}{2}x$$

Subtract \frac{1}{2}x from both sides.

$$\frac{1}{5}x-3-\frac{1}{2}x^{2}-\frac{1}{2}x=0$$

Combine \frac{1}{5}x and -\frac{1}{2}x to get -\frac{3}{10}x.

$$-\frac{3}{10}x-3-\frac{1}{2}x^{2}=0$$

Add 3 to both sides. Anything plus zero gives itself.

$$-\frac{3}{10}x-\frac{1}{2}x^{2}=3$$

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.

$$-\frac{1}{2}x^{2}-\frac{3}{10}x=3$$

Multiply both sides by -2.

$$\frac{-\frac{1}{2}x^{2}-\frac{3}{10}x}{-\frac{1}{2}}=\frac{3}{-\frac{1}{2}}$$

Dividing by -\frac{1}{2} undoes the multiplication by -\frac{1}{2}.

$$x^{2}+\left(-\frac{\frac{3}{10}}{-\frac{1}{2}}\right)x=\frac{3}{-\frac{1}{2}}$$

Divide -\frac{3}{10} by -\frac{1}{2} by multiplying -\frac{3}{10} by the reciprocal of -\frac{1}{2}.

$$x^{2}+\frac{3}{5}x=\frac{3}{-\frac{1}{2}}$$

Divide 3 by -\frac{1}{2} by multiplying 3 by the reciprocal of -\frac{1}{2}.

$$x^{2}+\frac{3}{5}x=-6$$

Divide \frac{3}{5}, the coefficient of the x term, by 2 to get \frac{3}{10}. Then add the square of \frac{3}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{3}{5}x+\left(\frac{3}{10}\right)^{2}=-6+\left(\frac{3}{10}\right)^{2}$$

Square \frac{3}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=-6+\frac{9}{100}$$

Add -6 to \frac{9}{100}.

$$x^{2}+\frac{3}{5}x+\frac{9}{100}=-\frac{591}{100}$$

Factor x^{2}+\frac{3}{5}x+\frac{9}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{3}{10}\right)^{2}=-\frac{591}{100}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{3}{10}\right)^{2}}=\sqrt{-\frac{591}{100}}$$

Simplify.

$$x+\frac{3}{10}=\frac{\sqrt{591}i}{10} x+\frac{3}{10}=-\frac{\sqrt{591}i}{10}$$

Subtract \frac{3}{10} from both sides of the equation.

$$x=\frac{-3+\sqrt{591}i}{10} x=\frac{-\sqrt{591}i-3}{10}$$