Solve for t
t = \frac{\sqrt{208401} + 51}{98} \approx 5.178669221
t=\frac{51-\sqrt{208401}}{98}\approx -4.137852895
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4.9t^{2}-5.1t=105
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.
4.9t^{2}-5.1t-105=105-105
Subtract 105 from both sides of the equation.
4.9t^{2}-5.1t-105=0
Subtracting 105 from itself leaves 0.
t=\frac{-\left(-5.1\right)±\sqrt{\left(-5.1\right)^{2}-4\times 4.9\left(-105\right)}}{2\times 4.9}
This equation is in standard form: ax^{2}+bx+c=0. Substitute 4.9 for a, -5.1 for b, and -105 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-\left(-5.1\right)±\sqrt{26.01-4\times 4.9\left(-105\right)}}{2\times 4.9}
Square -5.1 by squaring both the numerator and the denominator of the fraction.
t=\frac{-\left(-5.1\right)±\sqrt{26.01-19.6\left(-105\right)}}{2\times 4.9}
Multiply -4 times 4.9.
t=\frac{-\left(-5.1\right)±\sqrt{26.01+2058}}{2\times 4.9}
Multiply -19.6 times -105.
t=\frac{-\left(-5.1\right)±\sqrt{2084.01}}{2\times 4.9}
Add 26.01 to 2058.
t=\frac{-\left(-5.1\right)±\frac{\sqrt{208401}}{10}}{2\times 4.9}
Take the square root of 2084.01.
t=\frac{5.1±\frac{\sqrt{208401}}{10}}{2\times 4.9}
The opposite of -5.1 is 5.1.
t=\frac{5.1±\frac{\sqrt{208401}}{10}}{9.8}
Multiply 2 times 4.9.
t=\frac{\sqrt{208401}+51}{9.8\times 10}
Now solve the equation t=\frac{5.1±\frac{\sqrt{208401}}{10}}{9.8} when ± is plus. Add 5.1 to \frac{\sqrt{208401}}{10}.
t=\frac{\sqrt{208401}+51}{98}
Divide \frac{51+\sqrt{208401}}{10} by 9.8 by multiplying \frac{51+\sqrt{208401}}{10} by the reciprocal of 9.8.
t=\frac{51-\sqrt{208401}}{9.8\times 10}
Now solve the equation t=\frac{5.1±\frac{\sqrt{208401}}{10}}{9.8} when ± is minus. Subtract \frac{\sqrt{208401}}{10} from 5.1.
t=\frac{51-\sqrt{208401}}{98}
Divide \frac{51-\sqrt{208401}}{10} by 9.8 by multiplying \frac{51-\sqrt{208401}}{10} by the reciprocal of 9.8.
t=\frac{\sqrt{208401}+51}{98} t=\frac{51-\sqrt{208401}}{98}
The equation is now solved.
4.9t^{2}-5.1t=105
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.
\frac{4.9t^{2}-5.1t}{4.9}=\frac{105}{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}+\left(-\frac{5.1}{4.9}\right)t=\frac{105}{4.9}
Dividing by 4.9 undoes the multiplication by 4.9.
t^{2}-\frac{51}{49}t=\frac{105}{4.9}
Divide -5.1 by 4.9 by multiplying -5.1 by the reciprocal of 4.9.
t^{2}-\frac{51}{49}t=\frac{150}{7}
Divide 105 by 4.9 by multiplying 105 by the reciprocal of 4.9.
t^{2}-\frac{51}{49}t+\left(-\frac{51}{98}\right)^{2}=\frac{150}{7}+\left(-\frac{51}{98}\right)^{2}
Divide -\frac{51}{49}, the coefficient of the x term, by 2 to get -\frac{51}{98}. Then add the square of -\frac{51}{98} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{51}{49}t+\frac{2601}{9604}=\frac{150}{7}+\frac{2601}{9604}
Square -\frac{51}{98} by squaring both the numerator and the denominator of the fraction.
t^{2}-\frac{51}{49}t+\frac{2601}{9604}=\frac{208401}{9604}
Add \frac{150}{7} to \frac{2601}{9604} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{51}{98}\right)^{2}=\frac{208401}{9604}
Factor t^{2}-\frac{51}{49}t+\frac{2601}{9604}. 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{51}{98}\right)^{2}}=\sqrt{\frac{208401}{9604}}
Take the square root of both sides of the equation.
t-\frac{51}{98}=\frac{\sqrt{208401}}{98} t-\frac{51}{98}=-\frac{\sqrt{208401}}{98}
Simplify.
t=\frac{\sqrt{208401}+51}{98} t=\frac{51-\sqrt{208401}}{98}
Add \frac{51}{98} to both sides of the equation.
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