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5t^{2}-17t-7.25=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(-17\right)±\sqrt{\left(-17\right)^{2}-4\times 5\left(-7.25\right)}}{2\times 5}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -17 for b, and -7.25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-17\right)±\sqrt{289-4\times 5\left(-7.25\right)}}{2\times 5}
Square -17.
t=\frac{-\left(-17\right)±\sqrt{289-20\left(-7.25\right)}}{2\times 5}
Multiply -4 times 5.
t=\frac{-\left(-17\right)±\sqrt{289+145}}{2\times 5}
Multiply -20 times -7.25.
t=\frac{-\left(-17\right)±\sqrt{434}}{2\times 5}
Add 289 to 145.
t=\frac{17±\sqrt{434}}{2\times 5}
The opposite of -17 is 17.
t=\frac{17±\sqrt{434}}{10}
Multiply 2 times 5.
t=\frac{\sqrt{434}+17}{10}
Now solve the equation t=\frac{17±\sqrt{434}}{10} when ± is plus. Add 17 to \sqrt{434}.
t=\frac{17-\sqrt{434}}{10}
Now solve the equation t=\frac{17±\sqrt{434}}{10} when ± is minus. Subtract \sqrt{434} from 17.
t=\frac{\sqrt{434}+17}{10} t=\frac{17-\sqrt{434}}{10}
The equation is now solved.
5t^{2}-17t-7.25=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.
5t^{2}-17t-7.25-\left(-7.25\right)=-\left(-7.25\right)
Add 7.25 to both sides of the equation.
5t^{2}-17t=-\left(-7.25\right)
Subtracting -7.25 from itself leaves 0.
5t^{2}-17t=7.25
Subtract -7.25 from 0.
\frac{5t^{2}-17t}{5}=\frac{7.25}{5}
Divide both sides by 5.
t^{2}-\frac{17}{5}t=\frac{7.25}{5}
Dividing by 5 undoes the multiplication by 5.
t^{2}-\frac{17}{5}t=1.45
Divide 7.25 by 5.
t^{2}-\frac{17}{5}t+\left(-\frac{17}{10}\right)^{2}=1.45+\left(-\frac{17}{10}\right)^{2}
Divide -\frac{17}{5}, the coefficient of the x term, by 2 to get -\frac{17}{10}. Then add the square of -\frac{17}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{17}{5}t+\frac{289}{100}=1.45+\frac{289}{100}
Square -\frac{17}{10} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{17}{5}t+\frac{289}{100}=\frac{217}{50}
Add 1.45 to \frac{289}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{17}{10}\right)^{2}=\frac{217}{50}
Factor t^{2}-\frac{17}{5}t+\frac{289}{100}. 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{17}{10}\right)^{2}}=\sqrt{\frac{217}{50}}
Take the square root of both sides of the equation.
t-\frac{17}{10}=\frac{\sqrt{434}}{10} t-\frac{17}{10}=-\frac{\sqrt{434}}{10}
Simplify.
t=\frac{\sqrt{434}+17}{10} t=\frac{17-\sqrt{434}}{10}
Add \frac{17}{10} to both sides of the equation.