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