Solve x^2/2-3x/20=1/5

$$\frac { x ^ { 2 } } { 2 } - \frac { 3 x } { 20 } = \frac { 1 } { 5 }$$

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{4}{5}=0.8$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 20, the least common multiple of 2,20,5.

$$10x^{2}-3x=4$$

Subtract 4 from both sides.

$$10x^{2}-3x-4=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 10x^{2}+ax+bx-4. To find a and b, set up a system to be solved.

$$a+b=-3 ab=10\left(-4\right)=-40$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -40.

$$1,-40 2,-20 4,-10 5,-8$$

Calculate the sum for each pair.

$$1-40=-39 2-20=-18 4-10=-6 5-8=-3$$

The solution is the pair that gives sum -3.

$$a=-8 b=5$$

Rewrite 10x^{2}-3x-4 as \left(10x^{2}-8x\right)+\left(5x-4\right).

$$\left(10x^{2}-8x\right)+\left(5x-4\right)$$

Factor out 2x in 10x^{2}-8x.

$$2x\left(5x-4\right)+5x-4$$

Factor out common term 5x-4 by using distributive property.

$$\left(5x-4\right)\left(2x+1\right)$$

To find equation solutions, solve 5x-4=0 and 2x+1=0.

$$x=\frac{4}{5} x=-\frac{1}{2}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 20, the least common multiple of 2,20,5.

$$10x^{2}-3x=4$$

Subtract 4 from both sides.

$$10x^{2}-3x-4=0$$

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

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

Square -3.

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

Multiply -4 times 10.

$$x=\frac{-\left(-3\right)±\sqrt{9-40\left(-4\right)}}{2\times 10}$$

Multiply -40 times -4.

$$x=\frac{-\left(-3\right)±\sqrt{9+160}}{2\times 10}$$

Add 9 to 160.

$$x=\frac{-\left(-3\right)±\sqrt{169}}{2\times 10}$$

Take the square root of 169.

$$x=\frac{-\left(-3\right)±13}{2\times 10}$$

The opposite of -3 is 3.

$$x=\frac{3±13}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{3±13}{20}$$

Now solve the equation x=\frac{3±13}{20} when ± is plus. Add 3 to 13.

$$x=\frac{16}{20}$$

Reduce the fraction \frac{16}{20} to lowest terms by extracting and canceling out 4.

$$x=\frac{4}{5}$$

Now solve the equation x=\frac{3±13}{20} when ± is minus. Subtract 13 from 3.

$$x=-\frac{10}{20}$$

Reduce the fraction \frac{-10}{20} to lowest terms by extracting and canceling out 10.

$$x=-\frac{1}{2}$$

The equation is now solved.

$$x=\frac{4}{5} x=-\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 20, the least common multiple of 2,20,5.

$$10x^{2}-3x=4$$

Divide both sides by 10.

$$\frac{10x^{2}-3x}{10}=\frac{4}{10}$$

Dividing by 10 undoes the multiplication by 10.

$$x^{2}-\frac{3}{10}x=\frac{4}{10}$$

Reduce the fraction \frac{4}{10} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{3}{10}x=\frac{2}{5}$$

Divide -\frac{3}{10}, the coefficient of the x term, by 2 to get -\frac{3}{20}. Then add the square of -\frac{3}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{3}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=\frac{2}{5}+\frac{9}{400}$$

Add \frac{2}{5} to \frac{9}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{3}{10}x+\frac{9}{400}=\frac{169}{400}$$

Factor x^{2}-\frac{3}{10}x+\frac{9}{400}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{3}{20}\right)^{2}=\frac{169}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{3}{20}\right)^{2}}=\sqrt{\frac{169}{400}}$$

Simplify.

$$x-\frac{3}{20}=\frac{13}{20} x-\frac{3}{20}=-\frac{13}{20}$$

Add \frac{3}{20} to both sides of the equation.

$$x=\frac{4}{5} x=-\frac{1}{2}$$