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2x^{2}-11x-17=0
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=\frac{-\left(-11\right)±\sqrt{\left(-11\right)^{2}-4\times 2\left(-17\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -11 for b, and -17 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-11\right)±\sqrt{121-4\times 2\left(-17\right)}}{2\times 2}
Square -11.
x=\frac{-\left(-11\right)±\sqrt{121-8\left(-17\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-11\right)±\sqrt{121+136}}{2\times 2}
Multiply -8 times -17.
x=\frac{-\left(-11\right)±\sqrt{257}}{2\times 2}
Add 121 to 136.
x=\frac{11±\sqrt{257}}{2\times 2}
The opposite of -11 is 11.
x=\frac{11±\sqrt{257}}{4}
Multiply 2 times 2.
x=\frac{\sqrt{257}+11}{4}
Now solve the equation x=\frac{11±\sqrt{257}}{4} when ± is plus. Add 11 to \sqrt{257}.
x=\frac{11-\sqrt{257}}{4}
Now solve the equation x=\frac{11±\sqrt{257}}{4} when ± is minus. Subtract \sqrt{257} from 11.
x=\frac{\sqrt{257}+11}{4} x=\frac{11-\sqrt{257}}{4}
The equation is now solved.
2x^{2}-11x-17=0
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.
2x^{2}-11x-17-\left(-17\right)=-\left(-17\right)
Add 17 to both sides of the equation.
2x^{2}-11x=-\left(-17\right)
Subtracting -17 from itself leaves 0.
2x^{2}-11x=17
Subtract -17 from 0.
\frac{2x^{2}-11x}{2}=\frac{17}{2}
Divide both sides by 2.
x^{2}-\frac{11}{2}x=\frac{17}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=\frac{17}{2}+\left(-\frac{11}{4}\right)^{2}
Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{17}{2}+\frac{121}{16}
Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{257}{16}
Add \frac{17}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(x-\frac{11}{4}\right)^{2}=\frac{257}{16}
Factor x^{2}-\frac{11}{2}x+\frac{121}{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}.
\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{257}{16}}
Take the square root of both sides of the equation.
x-\frac{11}{4}=\frac{\sqrt{257}}{4} x-\frac{11}{4}=-\frac{\sqrt{257}}{4}
Simplify.
x=\frac{\sqrt{257}+11}{4} x=\frac{11-\sqrt{257}}{4}
Add \frac{11}{4} to both sides of the equation.