Solve for t
t = \frac{1000}{327} = 3\frac{19}{327} \approx 3.058103976
t=0
Share
Copied to clipboard
t\left(15-4.905t\right)=0
Factor out t.
t=0 t=\frac{1000}{327}
To find equation solutions, solve t=0 and 15-\frac{981t}{200}=0.
-4.905t^{2}+15t=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{-15±\sqrt{15^{2}}}{2\left(-4.905\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -4.905 for a, 15 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-15±15}{2\left(-4.905\right)}
Take the square root of 15^{2}.
t=\frac{-15±15}{-9.81}
Multiply 2 times -4.905.
t=\frac{0}{-9.81}
Now solve the equation t=\frac{-15±15}{-9.81} when ± is plus. Add -15 to 15.
t=0
Divide 0 by -9.81 by multiplying 0 by the reciprocal of -9.81.
t=-\frac{30}{-9.81}
Now solve the equation t=\frac{-15±15}{-9.81} when ± is minus. Subtract 15 from -15.
t=\frac{1000}{327}
Divide -30 by -9.81 by multiplying -30 by the reciprocal of -9.81.
t=0 t=\frac{1000}{327}
The equation is now solved.
-4.905t^{2}+15t=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.
\frac{-4.905t^{2}+15t}{-4.905}=\frac{0}{-4.905}
Divide both sides of the equation by -4.905, which is the same as multiplying both sides by the reciprocal of the fraction.
t^{2}+\frac{15}{-4.905}t=\frac{0}{-4.905}
Dividing by -4.905 undoes the multiplication by -4.905.
t^{2}-\frac{1000}{327}t=\frac{0}{-4.905}
Divide 15 by -4.905 by multiplying 15 by the reciprocal of -4.905.
t^{2}-\frac{1000}{327}t=0
Divide 0 by -4.905 by multiplying 0 by the reciprocal of -4.905.
t^{2}-\frac{1000}{327}t+\left(-\frac{500}{327}\right)^{2}=\left(-\frac{500}{327}\right)^{2}
Divide -\frac{1000}{327}, the coefficient of the x term, by 2 to get -\frac{500}{327}. Then add the square of -\frac{500}{327} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-\frac{1000}{327}t+\frac{250000}{106929}=\frac{250000}{106929}
Square -\frac{500}{327} by squaring both the numerator and the denominator of the fraction.
\left(t-\frac{500}{327}\right)^{2}=\frac{250000}{106929}
Factor t^{2}-\frac{1000}{327}t+\frac{250000}{106929}. 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{500}{327}\right)^{2}}=\sqrt{\frac{250000}{106929}}
Take the square root of both sides of the equation.
t-\frac{500}{327}=\frac{500}{327} t-\frac{500}{327}=-\frac{500}{327}
Simplify.
t=\frac{1000}{327} t=0
Add \frac{500}{327} to both sides of the equation.
Examples
Quadratic equation
{ x } ^ { 2 } - 4 x - 5 = 0
Trigonometry
4 \sin \theta \cos \theta = 2 \sin \theta
Linear equation
y = 3x + 4
Arithmetic
699 * 533
Matrix
\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]
Simultaneous equation
\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.
Differentiation
\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }
Integration
\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x
Limits
\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}