Solve 21/2frac{1}2^2+x=5*x^2

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

Solve for x

$$x=\frac{3\sqrt{6}}{20}+\frac{1}{10}\approx 0.467423461$$

$$x=-\frac{3\sqrt{6}}{20}+\frac{1}{10}\approx -0.267423461$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

Multiply 2 and 2 to get 4.

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

Add 4 and 1 to get 5.

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

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$5\times \frac{1}{4}+2x=10x^{2}$$

Multiply 5 and \frac{1}{4} to get \frac{5}{4}.

$$\frac{5}{4}+2x=10x^{2}$$

Subtract 10x^{2} from both sides.

$$\frac{5}{4}+2x-10x^{2}=0$$

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.

$$-10x^{2}+2x+\frac{5}{4}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -10 for a, 2 for b, and \frac{5}{4} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\left(-10\right)\times \frac{5}{4}}}{2\left(-10\right)}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\left(-10\right)\times \frac{5}{4}}}{2\left(-10\right)}$$

Multiply -4 times -10.

$$x=\frac{-2±\sqrt{4+40\times \frac{5}{4}}}{2\left(-10\right)}$$

Multiply 40 times \frac{5}{4}.

$$x=\frac{-2±\sqrt{4+50}}{2\left(-10\right)}$$

Add 4 to 50.

$$x=\frac{-2±\sqrt{54}}{2\left(-10\right)}$$

Take the square root of 54.

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

Multiply 2 times -10.

$$x=\frac{-2±3\sqrt{6}}{-20}$$

Now solve the equation x=\frac{-2±3\sqrt{6}}{-20} when ± is plus. Add -2 to 3\sqrt{6}.

$$x=\frac{3\sqrt{6}-2}{-20}$$

Divide -2+3\sqrt{6} by -20.

$$x=-\frac{3\sqrt{6}}{20}+\frac{1}{10}$$

Now solve the equation x=\frac{-2±3\sqrt{6}}{-20} when ± is minus. Subtract 3\sqrt{6} from -2.

$$x=\frac{-3\sqrt{6}-2}{-20}$$

Divide -2-3\sqrt{6} by -20.

$$x=\frac{3\sqrt{6}}{20}+\frac{1}{10}$$

The equation is now solved.

$$x=-\frac{3\sqrt{6}}{20}+\frac{1}{10} x=\frac{3\sqrt{6}}{20}+\frac{1}{10}$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

Multiply 2 and 2 to get 4.

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

Add 4 and 1 to get 5.

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

Calculate \frac{1}{2} to the power of 2 and get \frac{1}{4}.

$$5\times \frac{1}{4}+2x=10x^{2}$$

Multiply 5 and \frac{1}{4} to get \frac{5}{4}.

$$\frac{5}{4}+2x=10x^{2}$$

Subtract 10x^{2} from both sides.

$$\frac{5}{4}+2x-10x^{2}=0$$

Subtract \frac{5}{4} from both sides. Anything subtracted from zero gives its negation.

$$2x-10x^{2}=-\frac{5}{4}$$

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.

$$-10x^{2}+2x=-\frac{5}{4}$$

Divide both sides by -10.

$$\frac{-10x^{2}+2x}{-10}=-\frac{\frac{5}{4}}{-10}$$

Dividing by -10 undoes the multiplication by -10.

$$x^{2}+\frac{2}{-10}x=-\frac{\frac{5}{4}}{-10}$$

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

$$x^{2}-\frac{1}{5}x=-\frac{\frac{5}{4}}{-10}$$

Divide -\frac{5}{4} by -10.

$$x^{2}-\frac{1}{5}x=\frac{1}{8}$$

Divide -\frac{1}{5}, the coefficient of the x term, by 2 to get -\frac{1}{10}. Then add the square of -\frac{1}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{5}x+\left(-\frac{1}{10}\right)^{2}=\frac{1}{8}+\left(-\frac{1}{10}\right)^{2}$$

Square -\frac{1}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{1}{8}+\frac{1}{100}$$

Add \frac{1}{8} to \frac{1}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{5}x+\frac{1}{100}=\frac{27}{200}$$

Factor x^{2}-\frac{1}{5}x+\frac{1}{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}.

$$\left(x-\frac{1}{10}\right)^{2}=\frac{27}{200}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{10}\right)^{2}}=\sqrt{\frac{27}{200}}$$

Simplify.

$$x-\frac{1}{10}=\frac{3\sqrt{6}}{20} x-\frac{1}{10}=-\frac{3\sqrt{6}}{20}$$

Add \frac{1}{10} to both sides of the equation.

$$x=\frac{3\sqrt{6}}{20}+\frac{1}{10} x=-\frac{3\sqrt{6}}{20}+\frac{1}{10}$$