Solve 50x^2-25x+5=0

$$50 { x }^{ 2 } -25x+5 = 0$$

Solve for x (complex solution)

$$x=\frac{\sqrt{15}i}{20}+\frac{1}{4}\approx 0.25+0.193649167i$$

$$x=-\frac{\sqrt{15}i}{20}+\frac{1}{4}\approx 0.25-0.193649167i$$

Steps Using the Quadratic Formula

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.

$$50x^{2}-25x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 50 for a, -25 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-25\right)±\sqrt{\left(-25\right)^{2}-4\times 50\times 5}}{2\times 50}$$

Square -25.

$$x=\frac{-\left(-25\right)±\sqrt{625-4\times 50\times 5}}{2\times 50}$$

Multiply -4 times 50.

$$x=\frac{-\left(-25\right)±\sqrt{625-200\times 5}}{2\times 50}$$

Multiply -200 times 5.

$$x=\frac{-\left(-25\right)±\sqrt{625-1000}}{2\times 50}$$

Add 625 to -1000.

$$x=\frac{-\left(-25\right)±\sqrt{-375}}{2\times 50}$$

Take the square root of -375.

$$x=\frac{-\left(-25\right)±5\sqrt{15}i}{2\times 50}$$

The opposite of -25 is 25.

$$x=\frac{25±5\sqrt{15}i}{2\times 50}$$

Multiply 2 times 50.

$$x=\frac{25±5\sqrt{15}i}{100}$$

Now solve the equation x=\frac{25±5\sqrt{15}i}{100} when ± is plus. Add 25 to 5i\sqrt{15}.

$$x=\frac{25+5\sqrt{15}i}{100}$$

Divide 25+5i\sqrt{15} by 100.

$$x=\frac{\sqrt{15}i}{20}+\frac{1}{4}$$

Now solve the equation x=\frac{25±5\sqrt{15}i}{100} when ± is minus. Subtract 5i\sqrt{15} from 25.

$$x=\frac{-5\sqrt{15}i+25}{100}$$

Divide 25-5i\sqrt{15} by 100.

$$x=-\frac{\sqrt{15}i}{20}+\frac{1}{4}$$

The equation is now solved.

$$x=\frac{\sqrt{15}i}{20}+\frac{1}{4} x=-\frac{\sqrt{15}i}{20}+\frac{1}{4}$$

Steps for Completing the Square

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.

$$50x^{2}-25x+5=0$$

Subtract 5 from both sides of the equation.

$$50x^{2}-25x+5-5=-5$$

Subtracting 5 from itself leaves 0.

$$50x^{2}-25x=-5$$

Divide both sides by 50.

$$\frac{50x^{2}-25x}{50}=-\frac{5}{50}$$

Dividing by 50 undoes the multiplication by 50.

$$x^{2}+\left(-\frac{25}{50}\right)x=-\frac{5}{50}$$

Reduce the fraction \frac{-25}{50} to lowest terms by extracting and canceling out 25.

$$x^{2}-\frac{1}{2}x=-\frac{5}{50}$$

Reduce the fraction \frac{-5}{50} to lowest terms by extracting and canceling out 5.

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

Divide -\frac{1}{2}, the coefficient of the x term, by 2 to get -\frac{1}{4}. Then add the square of -\frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square -\frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{1}{10}+\frac{1}{16}$$

Add -\frac{1}{10} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{3}{80}$$

Factor x^{2}-\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{3}{80}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{3}{80}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{15}i}{20} x-\frac{1}{4}=-\frac{\sqrt{15}i}{20}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{\sqrt{15}i}{20}+\frac{1}{4} x=-\frac{\sqrt{15}i}{20}+\frac{1}{4}$$