Solve x^2-41x+800=0

$${ x }^{ 2 } -41x+800 = 0$$

Solve for x (complex solution)

$$x=\frac{41+7\sqrt{31}i}{2}\approx 20.5+19.48717527i$$

$$x=\frac{-7\sqrt{31}i+41}{2}\approx 20.5-19.48717527i$$

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.

$$x^{2}-41x+800=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -41 for b, and 800 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-41\right)±\sqrt{\left(-41\right)^{2}-4\times 800}}{2}$$

Square -41.

$$x=\frac{-\left(-41\right)±\sqrt{1681-4\times 800}}{2}$$

Multiply -4 times 800.

$$x=\frac{-\left(-41\right)±\sqrt{1681-3200}}{2}$$

Add 1681 to -3200.

$$x=\frac{-\left(-41\right)±\sqrt{-1519}}{2}$$

Take the square root of -1519.

$$x=\frac{-\left(-41\right)±7\sqrt{31}i}{2}$$

The opposite of -41 is 41.

$$x=\frac{41±7\sqrt{31}i}{2}$$

Now solve the equation x=\frac{41±7\sqrt{31}i}{2} when ± is plus. Add 41 to 7i\sqrt{31}.

$$x=\frac{41+7\sqrt{31}i}{2}$$

Now solve the equation x=\frac{41±7\sqrt{31}i}{2} when ± is minus. Subtract 7i\sqrt{31} from 41.

$$x=\frac{-7\sqrt{31}i+41}{2}$$

The equation is now solved.

$$x=\frac{41+7\sqrt{31}i}{2} x=\frac{-7\sqrt{31}i+41}{2}$$

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.

$$x^{2}-41x+800=0$$

Subtract 800 from both sides of the equation.

$$x^{2}-41x+800-800=-800$$

Subtracting 800 from itself leaves 0.

$$x^{2}-41x=-800$$

Divide -41, the coefficient of the x term, by 2 to get -\frac{41}{2}. Then add the square of -\frac{41}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-41x+\left(-\frac{41}{2}\right)^{2}=-800+\left(-\frac{41}{2}\right)^{2}$$

Square -\frac{41}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-41x+\frac{1681}{4}=-800+\frac{1681}{4}$$

Add -800 to \frac{1681}{4}.

$$x^{2}-41x+\frac{1681}{4}=-\frac{1519}{4}$$

Factor x^{2}-41x+\frac{1681}{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}.

$$\left(x-\frac{41}{2}\right)^{2}=-\frac{1519}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{41}{2}\right)^{2}}=\sqrt{-\frac{1519}{4}}$$

Simplify.

$$x-\frac{41}{2}=\frac{7\sqrt{31}i}{2} x-\frac{41}{2}=-\frac{7\sqrt{31}i}{2}$$

Add \frac{41}{2} to both sides of the equation.

$$x=\frac{41+7\sqrt{31}i}{2} x=\frac{-7\sqrt{31}i+41}{2}$$