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t^{2}+t+6.8=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.
t=\frac{-1±\sqrt{1^{2}-4\times 6.8}}{2}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 1 for b, and 6.8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-1±\sqrt{1-4\times 6.8}}{2}
Square 1.
t=\frac{-1±\sqrt{1-27.2}}{2}
Multiply -4 times 6.8.
t=\frac{-1±\sqrt{-26.2}}{2}
Add 1 to -27.2.
t=\frac{-1±\frac{\sqrt{655}i}{5}}{2}
Take the square root of -26.2.
t=\frac{\frac{\sqrt{655}i}{5}-1}{2}
Now solve the equation t=\frac{-1±\frac{\sqrt{655}i}{5}}{2} when ± is plus. Add -1 to \frac{i\sqrt{655}}{5}.
t=\frac{\sqrt{655}i}{10}-\frac{1}{2}
Divide -1+\frac{i\sqrt{655}}{5} by 2.
t=\frac{-\frac{\sqrt{655}i}{5}-1}{2}
Now solve the equation t=\frac{-1±\frac{\sqrt{655}i}{5}}{2} when ± is minus. Subtract \frac{i\sqrt{655}}{5} from -1.
t=-\frac{\sqrt{655}i}{10}-\frac{1}{2}
Divide -1-\frac{i\sqrt{655}}{5} by 2.
t=\frac{\sqrt{655}i}{10}-\frac{1}{2} t=-\frac{\sqrt{655}i}{10}-\frac{1}{2}
The equation is now solved.
t^{2}+t+6.8=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.
t^{2}+t+6.8-6.8=-6.8
Subtract 6.8 from both sides of the equation.
t^{2}+t=-6.8
Subtracting 6.8 from itself leaves 0.
t^{2}+t+\left(\frac{1}{2}\right)^{2}=-6.8+\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.
t^{2}+t+\frac{1}{4}=-6.8+\frac{1}{4}
Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.
t^{2}+t+\frac{1}{4}=-\frac{131}{20}
Add -6.8 to \frac{1}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t+\frac{1}{2}\right)^{2}=-\frac{131}{20}
Factor t^{2}+t+\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(t+\frac{1}{2}\right)^{2}}=\sqrt{-\frac{131}{20}}
Take the square root of both sides of the equation.
t+\frac{1}{2}=\frac{\sqrt{655}i}{10} t+\frac{1}{2}=-\frac{\sqrt{655}i}{10}
Simplify.
t=\frac{\sqrt{655}i}{10}-\frac{1}{2} t=-\frac{\sqrt{655}i}{10}-\frac{1}{2}
Subtract \frac{1}{2} from both sides of the equation.