Solve 2*x*x-17x+260=0

$$2 \times x \times x-17x+260 = 0$$

Solve for x (complex solution)

$$x=\frac{17+3\sqrt{199}i}{4}\approx 4.25+10.580051985i$$

$$x=\frac{-3\sqrt{199}i+17}{4}\approx 4.25-10.580051985i$$

Steps Using the Quadratic Formula

Multiply x and x to get x^{2}.

$$2x^{2}-17x+260=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -17 for b, and 260 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 2\times 260}}{2\times 2}$$

Square -17.

$$x=\frac{-\left(-17\right)±\sqrt{289-4\times 2\times 260}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-17\right)±\sqrt{289-8\times 260}}{2\times 2}$$

Multiply -8 times 260.

$$x=\frac{-\left(-17\right)±\sqrt{289-2080}}{2\times 2}$$

Add 289 to -2080.

$$x=\frac{-\left(-17\right)±\sqrt{-1791}}{2\times 2}$$

Take the square root of -1791.

$$x=\frac{-\left(-17\right)±3\sqrt{199}i}{2\times 2}$$

The opposite of -17 is 17.

$$x=\frac{17±3\sqrt{199}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{17±3\sqrt{199}i}{4}$$

Now solve the equation x=\frac{17±3\sqrt{199}i}{4} when ± is plus. Add 17 to 3i\sqrt{199}.

$$x=\frac{17+3\sqrt{199}i}{4}$$

Now solve the equation x=\frac{17±3\sqrt{199}i}{4} when ± is minus. Subtract 3i\sqrt{199} from 17.

$$x=\frac{-3\sqrt{199}i+17}{4}$$

The equation is now solved.

$$x=\frac{17+3\sqrt{199}i}{4} x=\frac{-3\sqrt{199}i+17}{4}$$

Steps for Completing the Square

Multiply x and x to get x^{2}.

$$2x^{2}-17x+260=0$$

Subtract 260 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}-17x=-260$$

Divide both sides by 2.

$$\frac{2x^{2}-17x}{2}=-\frac{260}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{17}{2}x=-\frac{260}{2}$$

Divide -260 by 2.

$$x^{2}-\frac{17}{2}x=-130$$

Divide -\frac{17}{2}, the coefficient of the x term, by 2 to get -\frac{17}{4}. Then add the square of -\frac{17}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{17}{2}x+\left(-\frac{17}{4}\right)^{2}=-130+\left(-\frac{17}{4}\right)^{2}$$

Square -\frac{17}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=-130+\frac{289}{16}$$

Add -130 to \frac{289}{16}.

$$x^{2}-\frac{17}{2}x+\frac{289}{16}=-\frac{1791}{16}$$

Factor x^{2}-\frac{17}{2}x+\frac{289}{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}.

$$\left(x-\frac{17}{4}\right)^{2}=-\frac{1791}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{17}{4}\right)^{2}}=\sqrt{-\frac{1791}{16}}$$

Simplify.

$$x-\frac{17}{4}=\frac{3\sqrt{199}i}{4} x-\frac{17}{4}=-\frac{3\sqrt{199}i}{4}$$

Add \frac{17}{4} to both sides of the equation.

$$x=\frac{17+3\sqrt{199}i}{4} x=\frac{-3\sqrt{199}i+17}{4}$$