Solve x^2-252929x+649212929=0

$${ x }^{ 2 } -252929x+649212929=0$$

Solve for x

$$x = \frac{5 \sqrt{2455049093} + 252929}{2} \approx 250335.629934501$$

$$x = \frac{252929 - 5 \sqrt{2455049093}}{2} \approx 2593.370065499$$

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.

$$x^{2}-252929x+649212929=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -252929 for b, and 649212929 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-252929\right)±\sqrt{\left(-252929\right)^{2}-4\times 649212929}}{2}$$

Square -252929.

$$x=\frac{-\left(-252929\right)±\sqrt{63973079041-4\times 649212929}}{2}$$

Multiply -4 times 649212929.

$$x=\frac{-\left(-252929\right)±\sqrt{63973079041-2596851716}}{2}$$

Add 63973079041 to -2596851716.

$$x=\frac{-\left(-252929\right)±\sqrt{61376227325}}{2}$$

Take the square root of 61376227325.

$$x=\frac{-\left(-252929\right)±5\sqrt{2455049093}}{2}$$

The opposite of -252929 is 252929.

$$x=\frac{252929±5\sqrt{2455049093}}{2}$$

Now solve the equation x=\frac{252929±5\sqrt{2455049093}}{2} when ± is plus. Add 252929 to 5\sqrt{2455049093}.

$$x=\frac{5\sqrt{2455049093}+252929}{2}$$

Now solve the equation x=\frac{252929±5\sqrt{2455049093}}{2} when ± is minus. Subtract 5\sqrt{2455049093} from 252929.

$$x=\frac{252929-5\sqrt{2455049093}}{2}$$

The equation is now solved.

$$x=\frac{5\sqrt{2455049093}+252929}{2} x=\frac{252929-5\sqrt{2455049093}}{2}$$

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.

$$x^{2}-252929x+649212929=0$$

Subtract 649212929 from both sides of the equation.

$$x^{2}-252929x+649212929-649212929=-649212929$$

Subtracting 649212929 from itself leaves 0.

$$x^{2}-252929x=-649212929$$

Divide -252929, the coefficient of the x term, by 2 to get -\frac{252929}{2}. Then add the square of -\frac{252929}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-252929x+\left(-\frac{252929}{2}\right)^{2}=-649212929+\left(-\frac{252929}{2}\right)^{2}$$

Square -\frac{252929}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-252929x+\frac{63973079041}{4}=-649212929+\frac{63973079041}{4}$$

Add -649212929 to \frac{63973079041}{4}.

$$x^{2}-252929x+\frac{63973079041}{4}=\frac{61376227325}{4}$$

Factor x^{2}-252929x+\frac{63973079041}{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}.

$$\left(x-\frac{252929}{2}\right)^{2}=\frac{61376227325}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{252929}{2}\right)^{2}}=\sqrt{\frac{61376227325}{4}}$$

Simplify.

$$x-\frac{252929}{2}=\frac{5\sqrt{2455049093}}{2} x-\frac{252929}{2}=-\frac{5\sqrt{2455049093}}{2}$$

Add \frac{252929}{2} to both sides of the equation.

$$x=\frac{5\sqrt{2455049093}+252929}{2} x=\frac{252929-5\sqrt{2455049093}}{2}$$