Solve 25841x^2+5487x+2584=0

$$25841 { x }^{ 2 } +5487x+2584=0$$

Solve for x (complex solution)

$$x=\frac{-5487+\sqrt{236985407}i}{51682}\approx -0.106168492+0.297866382i$$

$$x=\frac{-\sqrt{236985407}i-5487}{51682}\approx -0.106168492-0.297866382i$$

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.

$$25841x^{2}+5487x+2584=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25841 for a, 5487 for b, and 2584 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5487±\sqrt{5487^{2}-4\times 25841\times 2584}}{2\times 25841}$$

Square 5487.

$$x=\frac{-5487±\sqrt{30107169-4\times 25841\times 2584}}{2\times 25841}$$

Multiply -4 times 25841.

$$x=\frac{-5487±\sqrt{30107169-103364\times 2584}}{2\times 25841}$$

Multiply -103364 times 2584.

$$x=\frac{-5487±\sqrt{30107169-267092576}}{2\times 25841}$$

Add 30107169 to -267092576.

$$x=\frac{-5487±\sqrt{-236985407}}{2\times 25841}$$

Take the square root of -236985407.

$$x=\frac{-5487±\sqrt{236985407}i}{2\times 25841}$$

Multiply 2 times 25841.

$$x=\frac{-5487±\sqrt{236985407}i}{51682}$$

Now solve the equation x=\frac{-5487±\sqrt{236985407}i}{51682} when ± is plus. Add -5487 to i\sqrt{236985407}.

$$x=\frac{-5487+\sqrt{236985407}i}{51682}$$

Now solve the equation x=\frac{-5487±\sqrt{236985407}i}{51682} when ± is minus. Subtract i\sqrt{236985407} from -5487.

$$x=\frac{-\sqrt{236985407}i-5487}{51682}$$

The equation is now solved.

$$x=\frac{-5487+\sqrt{236985407}i}{51682} x=\frac{-\sqrt{236985407}i-5487}{51682}$$

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.

$$25841x^{2}+5487x+2584=0$$

Subtract 2584 from both sides of the equation.

$$25841x^{2}+5487x+2584-2584=-2584$$

Subtracting 2584 from itself leaves 0.

$$25841x^{2}+5487x=-2584$$

Divide both sides by 25841.

$$\frac{25841x^{2}+5487x}{25841}=-\frac{2584}{25841}$$

Dividing by 25841 undoes the multiplication by 25841.

$$x^{2}+\frac{5487}{25841}x=-\frac{2584}{25841}$$

Divide \frac{5487}{25841}, the coefficient of the x term, by 2 to get \frac{5487}{51682}. Then add the square of \frac{5487}{51682} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{5487}{25841}x+\left(\frac{5487}{51682}\right)^{2}=-\frac{2584}{25841}+\left(\frac{5487}{51682}\right)^{2}$$

Square \frac{5487}{51682} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{5487}{25841}x+\frac{30107169}{2671029124}=-\frac{2584}{25841}+\frac{30107169}{2671029124}$$

Add -\frac{2584}{25841} to \frac{30107169}{2671029124} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{5487}{25841}x+\frac{30107169}{2671029124}=-\frac{236985407}{2671029124}$$

Factor x^{2}+\frac{5487}{25841}x+\frac{30107169}{2671029124}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{5487}{51682}\right)^{2}=-\frac{236985407}{2671029124}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5487}{51682}\right)^{2}}=\sqrt{-\frac{236985407}{2671029124}}$$

Simplify.

$$x+\frac{5487}{51682}=\frac{\sqrt{236985407}i}{51682} x+\frac{5487}{51682}=-\frac{\sqrt{236985407}i}{51682}$$

Subtract \frac{5487}{51682} from both sides of the equation.

$$x=\frac{-5487+\sqrt{236985407}i}{51682} x=\frac{-\sqrt{236985407}i-5487}{51682}$$