Solve 100x^2=10x-1/4

$$100 x ^ { 2 } = 10 x - \frac { 1 } { 4 }$$

Solve for x

$$x=\frac{1}{20}=0.05$$

Steps Using the Quadratic Formula

Subtract 10x from both sides.

$$100x^{2}-10x=-\frac{1}{4}$$

Add \frac{1}{4} to both sides.

$$100x^{2}-10x+\frac{1}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 100 for a, -10 for b, and \frac{1}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Square -10.

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

Multiply -4 times 100.

$$x=\frac{-\left(-10\right)±\sqrt{100-400\times \frac{1}{4}}}{2\times 100}$$

Multiply -400 times \frac{1}{4}.

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

Add 100 to -100.

$$x=\frac{-\left(-10\right)±\sqrt{0}}{2\times 100}$$

Take the square root of 0.

$$x=-\frac{-10}{2\times 100}$$

The opposite of -10 is 10.

$$x=\frac{10}{2\times 100}$$

Multiply 2 times 100.

$$x=\frac{10}{200}$$

Reduce the fraction \frac{10}{200} to lowest terms by extracting and canceling out 10.

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

Steps for Completing the Square

Subtract 10x from both sides.

$$100x^{2}-10x=-\frac{1}{4}$$

Divide both sides by 100.

$$\frac{100x^{2}-10x}{100}=-\frac{\frac{1}{4}}{100}$$

Dividing by 100 undoes the multiplication by 100.

$$x^{2}+\left(-\frac{10}{100}\right)x=-\frac{\frac{1}{4}}{100}$$

Reduce the fraction \frac{-10}{100} to lowest terms by extracting and canceling out 10.

$$x^{2}-\frac{1}{10}x=-\frac{\frac{1}{4}}{100}$$

Divide -\frac{1}{4} by 100.

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

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.

$$x^{2}-\frac{1}{10}x+\left(-\frac{1}{20}\right)^{2}=-\frac{1}{400}+\left(-\frac{1}{20}\right)^{2}$$

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

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

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

$$x^{2}-\frac{1}{10}x+\frac{1}{400}=0$$

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}.

$$\left(x-\frac{1}{20}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{20}\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-\frac{1}{20}=0 x-\frac{1}{20}=0$$

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

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

The equation is now solved. Solutions are the same.

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