Solve 1=2(x^2-5x+25)

$$1 = 2 ( x ^ { 2 } - 5 x + 25 )$$

Solve for x (complex solution)

$$x=\frac{5+\sqrt{73}i}{2}\approx 2.5+4.272001873i$$

$$x=\frac{-\sqrt{73}i+5}{2}\approx 2.5-4.272001873i$$

Steps Using the Quadratic Formula

Divide both sides by 2.

$$\frac{1}{2}=x^{2}-5x+25$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-5x+25=\frac{1}{2}$$

Subtract \frac{1}{2} from both sides.

$$x^{2}-5x+25-\frac{1}{2}=0$$

Subtract \frac{1}{2} from 25 to get \frac{49}{2}.

$$x^{2}-5x+\frac{49}{2}=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -5 for b, and \frac{49}{2} for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times \frac{49}{2}}}{2}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times \frac{49}{2}}}{2}$$

Multiply -4 times \frac{49}{2}.

$$x=\frac{-\left(-5\right)±\sqrt{25-98}}{2}$$

Add 25 to -98.

$$x=\frac{-\left(-5\right)±\sqrt{-73}}{2}$$

Take the square root of -73.

$$x=\frac{-\left(-5\right)±\sqrt{73}i}{2}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{73}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{73}i}{2} when ± is plus. Add 5 to i\sqrt{73}.

$$x=\frac{5+\sqrt{73}i}{2}$$

Now solve the equation x=\frac{5±\sqrt{73}i}{2} when ± is minus. Subtract i\sqrt{73} from 5.

$$x=\frac{-\sqrt{73}i+5}{2}$$

The equation is now solved.

$$x=\frac{5+\sqrt{73}i}{2} x=\frac{-\sqrt{73}i+5}{2}$$

Steps for Completing the Square

Divide both sides by 2.

$$\frac{1}{2}=x^{2}-5x+25$$

Swap sides so that all variable terms are on the left hand side.

$$x^{2}-5x+25=\frac{1}{2}$$

Subtract 25 from both sides.

$$x^{2}-5x=\frac{1}{2}-25$$

Subtract 25 from \frac{1}{2} to get -\frac{49}{2}.

$$x^{2}-5x=-\frac{49}{2}$$

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

$$x^{2}-5x+\left(-\frac{5}{2}\right)^{2}=-\frac{49}{2}+\left(-\frac{5}{2}\right)^{2}$$

Square -\frac{5}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-5x+\frac{25}{4}=-\frac{49}{2}+\frac{25}{4}$$

Add -\frac{49}{2} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-5x+\frac{25}{4}=-\frac{73}{4}$$

Factor x^{2}-5x+\frac{25}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{2}\right)^{2}=-\frac{73}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{2}\right)^{2}}=\sqrt{-\frac{73}{4}}$$

Simplify.

$$x-\frac{5}{2}=\frac{\sqrt{73}i}{2} x-\frac{5}{2}=-\frac{\sqrt{73}i}{2}$$

Add \frac{5}{2} to both sides of the equation.

$$x=\frac{5+\sqrt{73}i}{2} x=\frac{-\sqrt{73}i+5}{2}$$