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2x^{2}-5x-10=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(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 2\left(-10\right)}}{2\times 2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -5 for b, and -10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-\left(-5\right)±\sqrt{25-4\times 2\left(-10\right)}}{2\times 2}
Square -5.
x=\frac{-\left(-5\right)±\sqrt{25-8\left(-10\right)}}{2\times 2}
Multiply -4 times 2.
x=\frac{-\left(-5\right)±\sqrt{25+80}}{2\times 2}
Multiply -8 times -10.
x=\frac{-\left(-5\right)±\sqrt{105}}{2\times 2}
Add 25 to 80.
x=\frac{5±\sqrt{105}}{2\times 2}
The opposite of -5 is 5.
x=\frac{5±\sqrt{105}}{4}
Multiply 2 times 2.
x=\frac{\sqrt{105}+5}{4}
Now solve the equation x=\frac{5±\sqrt{105}}{4} when ± is plus. Add 5 to \sqrt{105}.
x=\frac{5-\sqrt{105}}{4}
Now solve the equation x=\frac{5±\sqrt{105}}{4} when ± is minus. Subtract \sqrt{105} from 5.
x=\frac{\sqrt{105}+5}{4} x=\frac{5-\sqrt{105}}{4}
The equation is now solved.
2x^{2}-5x-10=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}-5x-10-\left(-10\right)=-\left(-10\right)
Add 10 to both sides of the equation.
2x^{2}-5x=-\left(-10\right)
Subtracting -10 from itself leaves 0.
2x^{2}-5x=10
Subtract -10 from 0.
\frac{2x^{2}-5x}{2}=\frac{10}{2}
Divide both sides by 2.
x^{2}-\frac{5}{2}x=\frac{10}{2}
Dividing by 2 undoes the multiplication by 2.
x^{2}-\frac{5}{2}x=5
Divide 10 by 2.
x^{2}-\frac{5}{2}x+\left(-\frac{5}{4}\right)^{2}=5+\left(-\frac{5}{4}\right)^{2}
Divide -\frac{5}{2}, the coefficient of the x term, by 2 to get -\frac{5}{4}. Then add the square of -\frac{5}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}-\frac{5}{2}x+\frac{25}{16}=5+\frac{25}{16}
Square -\frac{5}{4} by squaring both the numerator and the denominator of the fraction.
x^{2}-\frac{5}{2}x+\frac{25}{16}=\frac{105}{16}
Add 5 to \frac{25}{16}.
\left(x-\frac{5}{4}\right)^{2}=\frac{105}{16}
Factor x^{2}-\frac{5}{2}x+\frac{25}{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{5}{4}\right)^{2}}=\sqrt{\frac{105}{16}}
Take the square root of both sides of the equation.
x-\frac{5}{4}=\frac{\sqrt{105}}{4} x-\frac{5}{4}=-\frac{\sqrt{105}}{4}
Simplify.
x=\frac{\sqrt{105}+5}{4} x=\frac{5-\sqrt{105}}{4}
Add \frac{5}{4} to both sides of the equation.