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-4,9t^{2}+29,4t-33,05=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{-29,4±\sqrt{29,4^{2}-4\left(-4,9\right)\left(-33,05\right)}}{2\left(-4,9\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -4,9 for a, 29,4 for b, and -33,05 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-29,4±\sqrt{864,36-4\left(-4,9\right)\left(-33,05\right)}}{2\left(-4,9\right)}
Square 29,4 by squaring both the numerator and the denominator of the fraction.
t=\frac{-29,4±\sqrt{864,36+19,6\left(-33,05\right)}}{2\left(-4,9\right)}
Multiply -4 times -4,9.
t=\frac{-29,4±\sqrt{864,36-647,78}}{2\left(-4,9\right)}
Multiply 19,6 times -33,05 by multiplying numerator times numerator and denominator times denominator. Then reduce the fraction to lowest terms if possible.
t=\frac{-29,4±\sqrt{216,58}}{2\left(-4,9\right)}
Add 864,36 to -647,78 by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
t=\frac{-29,4±\frac{7\sqrt{442}}{10}}{2\left(-4,9\right)}
Take the square root of 216,58.
t=\frac{-29,4±\frac{7\sqrt{442}}{10}}{-9,8}
Multiply 2 times -4,9.
t=\frac{\frac{7\sqrt{442}}{10}-\frac{147}{5}}{-9,8}
Now solve the equation t=\frac{-29,4±\frac{7\sqrt{442}}{10}}{-9,8} when ± is plus. Add -29,4 to \frac{7\sqrt{442}}{10}.
t=-\frac{\sqrt{442}}{14}+3
Divide -\frac{147}{5}+\frac{7\sqrt{442}}{10} by -9,8 by multiplying -\frac{147}{5}+\frac{7\sqrt{442}}{10} by the reciprocal of -9,8.
t=\frac{-\frac{7\sqrt{442}}{10}-\frac{147}{5}}{-9,8}
Now solve the equation t=\frac{-29,4±\frac{7\sqrt{442}}{10}}{-9,8} when ± is minus. Subtract \frac{7\sqrt{442}}{10} from -29,4.
t=\frac{\sqrt{442}}{14}+3
Divide -\frac{147}{5}-\frac{7\sqrt{442}}{10} by -9,8 by multiplying -\frac{147}{5}-\frac{7\sqrt{442}}{10} by the reciprocal of -9,8.
t=-\frac{\sqrt{442}}{14}+3 t=\frac{\sqrt{442}}{14}+3
The equation is now solved.
-4,9t^{2}+29,4t-33,05=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.
-4,9t^{2}+29,4t-33,05-\left(-33,05\right)=-\left(-33,05\right)
Add 33,05 to both sides of the equation.
-4,9t^{2}+29,4t=-\left(-33,05\right)
Subtracting -33,05 from itself leaves 0.
-4,9t^{2}+29,4t=33,05
Subtract -33,05 from 0.
\frac{-4,9t^{2}+29,4t}{-4,9}=\frac{33,05}{-4,9}
Divide both sides of the equation by -4,9, which is the same as multiplying both sides by the reciprocal of the fraction.
t^{2}+\frac{29,4}{-4,9}t=\frac{33,05}{-4,9}
Dividing by -4,9 undoes the multiplication by -4,9.
t^{2}-6t=\frac{33,05}{-4,9}
Divide 29,4 by -4,9 by multiplying 29,4 by the reciprocal of -4,9.
t^{2}-6t=-\frac{661}{98}
Divide 33,05 by -4,9 by multiplying 33,05 by the reciprocal of -4,9.
t^{2}-6t+\left(-3\right)^{2}=-\frac{661}{98}+\left(-3\right)^{2}
Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-6t+9=-\frac{661}{98}+9
Square -3.
t^{2}-6t+9=\frac{221}{98}
Add -\frac{661}{98} to 9.
\left(t-3\right)^{2}=\frac{221}{98}
Factor t^{2}-6t+9. 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-3\right)^{2}}=\sqrt{\frac{221}{98}}
Take the square root of both sides of the equation.
t-3=\frac{\sqrt{442}}{14} t-3=-\frac{\sqrt{442}}{14}
Simplify.
t=\frac{\sqrt{442}}{14}+3 t=-\frac{\sqrt{442}}{14}+3
Add 3 to both sides of the equation.