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x^{2}+5x+1=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{-5±\sqrt{5^{2}-4}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 5 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
x=\frac{-5±\sqrt{25-4}}{2}
Square 5.
x=\frac{-5±\sqrt{21}}{2}
Add 25 to -4.
x=\frac{\sqrt{21}-5}{2}
Now solve the equation x=\frac{-5±\sqrt{21}}{2} when ± is plus. Add -5 to \sqrt{21}.
x=\frac{-\sqrt{21}-5}{2}
Now solve the equation x=\frac{-5±\sqrt{21}}{2} when ± is minus. Subtract \sqrt{21} from -5.
x=\frac{\sqrt{21}-5}{2} x=\frac{-\sqrt{21}-5}{2}
The equation is now solved.
x^{2}+5x+1=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.
x^{2}+5x+1-1=-1
Subtract 1 from both sides of the equation.
x^{2}+5x=-1
Subtracting 1 from itself leaves 0.
x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-1+\left(\frac{5}{2}\right)^{2}
Divide 5, the coefficient of the x term, by 2 to get \frac{5}{2}. Then add the square of \frac{5}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
x^{2}+5x+\frac{25}{4}=-1+\frac{25}{4}
Square \frac{5}{2} by squaring both the numerator and the denominator of the fraction.
x^{2}+5x+\frac{25}{4}=\frac{21}{4}
Add -1 to \frac{25}{4}.
\left(x+\frac{5}{2}\right)^{2}=\frac{21}{4}
Factor x^{2}+5x+\frac{25}{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}.
\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{21}{4}}
Take the square root of both sides of the equation.
x+\frac{5}{2}=\frac{\sqrt{21}}{2} x+\frac{5}{2}=-\frac{\sqrt{21}}{2}
Simplify.
x=\frac{\sqrt{21}-5}{2} x=\frac{-\sqrt{21}-5}{2}
Subtract \frac{5}{2} from both sides of the equation.