Solve 10x^2=x+2

$$10 x ^ { 2 } = x + 2$$

Solve for x

$$x=-\frac{2}{5}=-0.4$$

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

Steps Using Factoring By Grouping

Subtract x from both sides.

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

Subtract 2 from both sides.

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

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

$$a+b=-1 ab=10\left(-2\right)=-20$$

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 -20.

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

Calculate the sum for each pair.

$$1-20=-19 2-10=-8 4-5=-1$$

The solution is the pair that gives sum -1.

$$a=-5 b=4$$

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

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

Factor out 5x in the first and 2 in the second group.

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

Factor out common term 2x-1 by using distributive property.

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

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

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

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Subtract 2 from both sides.

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

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

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

Multiply -4 times 10.

$$x=\frac{-\left(-1\right)±\sqrt{1-40\left(-2\right)}}{2\times 10}$$

Multiply -40 times -2.

$$x=\frac{-\left(-1\right)±\sqrt{1+80}}{2\times 10}$$

Add 1 to 80.

$$x=\frac{-\left(-1\right)±\sqrt{81}}{2\times 10}$$

Take the square root of 81.

$$x=\frac{-\left(-1\right)±9}{2\times 10}$$

The opposite of -1 is 1.

$$x=\frac{1±9}{2\times 10}$$

Multiply 2 times 10.

$$x=\frac{1±9}{20}$$

Now solve the equation x=\frac{1±9}{20} when ± is plus. Add 1 to 9.

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

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

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

Now solve the equation x=\frac{1±9}{20} when ± is minus. Subtract 9 from 1.

$$x=-\frac{8}{20}$$

Reduce the fraction \frac{-8}{20} to lowest terms by extracting and canceling out 4.

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

The equation is now solved.

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

Steps for Completing the Square

Subtract x from both sides.

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

Divide both sides by 10.

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

Dividing by 10 undoes the multiplication by 10.

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

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

$$x^{2}-\frac{1}{10}x=\frac{1}{5}$$

Divide -\frac{1}{10}, the coefficient of the x term, by 2 to get -\frac{1}{20}. Then add the square of -\frac{1}{20} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=\frac{1}{5}+\left(-\frac{1}{20}\right)^{2}$$

Square -\frac{1}{20} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{1}{5}+\frac{1}{400}$$

Add \frac{1}{5} to \frac{1}{400} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=\frac{81}{400}$$

Factor x^{2}-\frac{1}{10}x+\frac{1}{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}.

$$\left(x-\frac{1}{20}\right)^{2}=\frac{81}{400}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{\frac{81}{400}}$$

Simplify.

$$x-\frac{1}{20}=\frac{9}{20} x-\frac{1}{20}=-\frac{9}{20}$$

Add \frac{1}{20} to both sides of the equation.

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