Solve for t
t = \frac{\sqrt{521} - 1}{13} \approx 1.678878802
t=\frac{-\sqrt{521}-1}{13}\approx -1.832724955
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-3.25t^{2}-0.5t+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.
t=\frac{-\left(-0.5\right)±\sqrt{\left(-0.5\right)^{2}-4\left(-3.25\right)\times 10}}{2\left(-3.25\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -3.25 for a, -0.5 for b, and 10 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-0.5\right)±\sqrt{0.25-4\left(-3.25\right)\times 10}}{2\left(-3.25\right)}
Square -0.5 by squaring both the numerator and the denominator of the fraction.
t=\frac{-\left(-0.5\right)±\sqrt{0.25+13\times 10}}{2\left(-3.25\right)}
Multiply -4 times -3.25.
t=\frac{-\left(-0.5\right)±\sqrt{0.25+130}}{2\left(-3.25\right)}
Multiply 13 times 10.
t=\frac{-\left(-0.5\right)±\sqrt{130.25}}{2\left(-3.25\right)}
Add 0.25 to 130.
t=\frac{-\left(-0.5\right)±\frac{\sqrt{521}}{2}}{2\left(-3.25\right)}
Take the square root of 130.25.
t=\frac{0.5±\frac{\sqrt{521}}{2}}{2\left(-3.25\right)}
The opposite of -0.5 is 0.5.
t=\frac{0.5±\frac{\sqrt{521}}{2}}{-6.5}
Multiply 2 times -3.25.
t=\frac{\sqrt{521}+1}{-6.5\times 2}
Now solve the equation t=\frac{0.5±\frac{\sqrt{521}}{2}}{-6.5} when ± is plus. Add 0.5 to \frac{\sqrt{521}}{2}.
t=\frac{-\sqrt{521}-1}{13}
Divide \frac{1+\sqrt{521}}{2} by -6.5 by multiplying \frac{1+\sqrt{521}}{2} by the reciprocal of -6.5.
t=\frac{1-\sqrt{521}}{-6.5\times 2}
Now solve the equation t=\frac{0.5±\frac{\sqrt{521}}{2}}{-6.5} when ± is minus. Subtract \frac{\sqrt{521}}{2} from 0.5.
t=\frac{\sqrt{521}-1}{13}
Divide \frac{1-\sqrt{521}}{2} by -6.5 by multiplying \frac{1-\sqrt{521}}{2} by the reciprocal of -6.5.
t=\frac{-\sqrt{521}-1}{13} t=\frac{\sqrt{521}-1}{13}
The equation is now solved.
-3.25t^{2}-0.5t+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.
-3.25t^{2}-0.5t+10-10=-10
Subtract 10 from both sides of the equation.
-3.25t^{2}-0.5t=-10
Subtracting 10 from itself leaves 0.
\frac{-3.25t^{2}-0.5t}{-3.25}=-\frac{10}{-3.25}
Divide both sides of the equation by -3.25, which is the same as multiplying both sides by the reciprocal of the fraction.
t^{2}+\left(-\frac{0.5}{-3.25}\right)t=-\frac{10}{-3.25}
Dividing by -3.25 undoes the multiplication by -3.25.
t^{2}+\frac{2}{13}t=-\frac{10}{-3.25}
Divide -0.5 by -3.25 by multiplying -0.5 by the reciprocal of -3.25.
t^{2}+\frac{2}{13}t=\frac{40}{13}
Divide -10 by -3.25 by multiplying -10 by the reciprocal of -3.25.
t^{2}+\frac{2}{13}t+\frac{1}{13}^{2}=\frac{40}{13}+\frac{1}{13}^{2}
Divide \frac{2}{13}, the coefficient of the x term, by 2 to get \frac{1}{13}. Then add the square of \frac{1}{13} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}+\frac{2}{13}t+\frac{1}{169}=\frac{40}{13}+\frac{1}{169}
Square \frac{1}{13} by squaring both the numerator and the denominator of the fraction.
t^{2}+\frac{2}{13}t+\frac{1}{169}=\frac{521}{169}
Add \frac{40}{13} to \frac{1}{169} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t+\frac{1}{13}\right)^{2}=\frac{521}{169}
Factor t^{2}+\frac{2}{13}t+\frac{1}{169}. 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}{13}\right)^{2}}=\sqrt{\frac{521}{169}}
Take the square root of both sides of the equation.
t+\frac{1}{13}=\frac{\sqrt{521}}{13} t+\frac{1}{13}=-\frac{\sqrt{521}}{13}
Simplify.
t=\frac{\sqrt{521}-1}{13} t=\frac{-\sqrt{521}-1}{13}
Subtract \frac{1}{13} from both sides of the equation.
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Limits
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