Solve for t
t = \frac{\sqrt{106} + 9}{2} \approx 9.64781507
t=\frac{9-\sqrt{106}}{2}\approx -0.64781507
Share
Copied to clipboard
-16t^{2}+144t+100=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{-144±\sqrt{144^{2}-4\left(-16\right)\times 100}}{2\left(-16\right)}
This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, 144 for b, and 100 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.
t=\frac{-144±\sqrt{20736-4\left(-16\right)\times 100}}{2\left(-16\right)}
Square 144.
t=\frac{-144±\sqrt{20736+64\times 100}}{2\left(-16\right)}
Multiply -4 times -16.
t=\frac{-144±\sqrt{20736+6400}}{2\left(-16\right)}
Multiply 64 times 100.
t=\frac{-144±\sqrt{27136}}{2\left(-16\right)}
Add 20736 to 6400.
t=\frac{-144±16\sqrt{106}}{2\left(-16\right)}
Take the square root of 27136.
t=\frac{-144±16\sqrt{106}}{-32}
Multiply 2 times -16.
t=\frac{16\sqrt{106}-144}{-32}
Now solve the equation t=\frac{-144±16\sqrt{106}}{-32} when ± is plus. Add -144 to 16\sqrt{106}.
t=\frac{9-\sqrt{106}}{2}
Divide -144+16\sqrt{106} by -32.
t=\frac{-16\sqrt{106}-144}{-32}
Now solve the equation t=\frac{-144±16\sqrt{106}}{-32} when ± is minus. Subtract 16\sqrt{106} from -144.
t=\frac{\sqrt{106}+9}{2}
Divide -144-16\sqrt{106} by -32.
t=\frac{9-\sqrt{106}}{2} t=\frac{\sqrt{106}+9}{2}
The equation is now solved.
-16t^{2}+144t+100=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.
-16t^{2}+144t+100-100=-100
Subtract 100 from both sides of the equation.
-16t^{2}+144t=-100
Subtracting 100 from itself leaves 0.
\frac{-16t^{2}+144t}{-16}=-\frac{100}{-16}
Divide both sides by -16.
t^{2}+\frac{144}{-16}t=-\frac{100}{-16}
Dividing by -16 undoes the multiplication by -16.
t^{2}-9t=-\frac{100}{-16}
Divide 144 by -16.
t^{2}-9t=\frac{25}{4}
Reduce the fraction \frac{-100}{-16} to lowest terms by extracting and canceling out 4.
t^{2}-9t+\left(-\frac{9}{2}\right)^{2}=\frac{25}{4}+\left(-\frac{9}{2}\right)^{2}
Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.
t^{2}-9t+\frac{81}{4}=\frac{25+81}{4}
Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.
t^{2}-9t+\frac{81}{4}=\frac{53}{2}
Add \frac{25}{4} to \frac{81}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.
\left(t-\frac{9}{2}\right)^{2}=\frac{53}{2}
Factor t^{2}-9t+\frac{81}{4}. 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{9}{2}\right)^{2}}=\sqrt{\frac{53}{2}}
Take the square root of both sides of the equation.
t-\frac{9}{2}=\frac{\sqrt{106}}{2} t-\frac{9}{2}=-\frac{\sqrt{106}}{2}
Simplify.
t=\frac{\sqrt{106}+9}{2} t=\frac{9-\sqrt{106}}{2}
Add \frac{9}{2} 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}